Number systems: CAT Previous Year Questions

Ans: 6

Based on the problem statement, the 3-digit number N comprises non-zero and distinct digits.
The available digits for forming this 3-digit number are therefore: $2,3,5,6,7,8$
The problem further specifies that precisely one of these digits is a prime number.
To find the smallest possible value of N, we select the smallest prime digit and arrange the remaining non-prime digits in ascending order. The smallest prime digit from the given set is 2. The remaining non-prime digits are 6 and 8. Arranging these in ascending order gives us the minimum possible value for N as 268.
Now, let’s determine the prime factorization of 268:
$268 = 4 \times 67 = 2^2 \times 67^1$
The total count of factors for a number with prime factorization $p_1^{a_1} p_2^{a_2} \dots p_k^{a_k}$ is given by $ (a_1+1)(a_2+1)\dots(a_k+1) $.
Applying this formula to 268:
Number of factors = $ (2+1)(1+1) = 3 \times 2 = 6 $

Ans: A

For values of n equal to 3, 4, and 5, and for x within the interval [n, n+1), the floor of x is n, i.e., $\lfloor x \rfloor = n$. Consequently, the given equation simplifies to:
$\lfloor x^2 \rfloor = \lfloor x \rfloor^2 = n^2$
This implies that $x^2$ must fall within the interval $[n^2, n^2+1)$. Taking the square root of these bounds, we find that x must be in the interval $[n, \sqrt{n^2+1})$.

Therefore, for the range $3 \le x \le 6$, the set of possible x values (S) is the union of the following intervals:
$S = [3, \sqrt{3^2+1}) \cup [4, \sqrt{4^2+1}) \cup [5, \sqrt{5^2+1}) \cup \{6\}$
$S = [3, \sqrt{10}) \cup [4, \sqrt{17}) \cup [5, \sqrt{26}) \cup \{6\}$

Now let’s evaluate the given options:
Options B and C include $\sqrt{10}$, which is not contained within the interval $[3, \sqrt{10})$.
Option D includes $\sqrt{18}$, which is outside the calculated range.
Thus, options B, C, and D are not correct.

Option A is the only remaining and therefore the correct answer.

Ans: 25

Explanation:Let the given number be N.
$N = (625)^{65} \times (128)^{36}$
We can rewrite the bases in terms of their prime factors:
$625 = 5^4$
$128 = 2^7$
Substituting these into the expression for N:
$N = (5^4)^{65} \times (2^7)^{36}$
Using the exponent rule $(a^m)^n = a^{m \times n}$:
$N = 5^{4 \times 65} \times 2^{7 \times 36}$
$N = 5^{260} \times 2^{252}$
To find the sum of the digits, we need to express N in the form $a \times 10^k$. We can do this by pairing up powers of 2 and 5 to form powers of 10.
$N = 5^{260} \times 2^{252} = 5^{252} \times 5^{260-252} \times 2^{252}$
$N = 5^{252} \times 5^8 \times 2^{252}$
$N = (5 \times 2)^{252} \times 5^8$
$N = 10^{252} \times 5^8$
Now, let’s calculate $5^8$:
$5^1 = 5$
$5^2 = 25$
$5^3 = 125$
$5^4 = 625$
$5^5 = 3125$
$5^6 = 15625$
$5^7 = 78125$
$5^8 = 390625$
So, N can be written as:
$N = 390625 \times 10^{252}$
This means that the number N is 390625 followed by 252 zeros.
The digits of N are 3, 9, 0, 6, 2, 5, followed by 252 zeros.
The sum of the digits of N is the sum of these non-zero digits plus the sum of the zeros.
Sum of digits = 3 + 9 + 0 + 6 + 2 + 5 + (252 * 0)
Sum of digits = 3 + 9 + 6 + 2 + 5
Sum of digits = 12 + 6 + 2 + 5
Sum of digits = 18 + 2 + 5
Sum of digits = 20 + 5
Sum of digits = 25

Final_Answer:25

Ans: D

Any factor of the given number will be of the structure $2^a*3^b*5^c*7^d$ where the exponents satisfy $0\le a\le6,\ 0\le b\le5,\ 0\le c\le3,\ 0\le d\le2$.
Since the factors must be of the form 3r+1, they cannot be multiples of 3, which implies b must be 0.
Let’s consider the remainders when the prime factors are divided by 3:
$2 \pmod 3 = 2$
$5 \pmod 3 = 2$
$7 \pmod 3 = 1$
Therefore, a factor of the form $2^a5^c7^d$ will have a remainder modulo 3 as:
$2^a5^c7^d \equiv 2^a \cdot 2^c \cdot 1^d \pmod 3$
$ \equiv 2^{a+c} \pmod 3$
For $2^k$ to be of the form 3r+1, k must be an even number. Thus, we require the sum of the exponents a+c to be even.
The possible values for ‘a’ are $0, 1, 2, 3, 4, 5, 6$. Among these, there are 4 even values (0, 2, 4, 6) and 3 odd values (1, 3, 5).
The possible values for ‘c’ are $0, 1, 2, 3$. Among these, there are 2 even values (0, 2) and 2 odd values (1, 3).
The number of pairs (a,c) such that (a+c) is even occurs in two scenarios:
1. ‘a’ is even and ‘c’ is even: $4 \text{ (choices for a)} \times 2 \text{ (choices for c)} = 8$
2. ‘a’ is odd and ‘c’ is odd: $3 \text{ (choices for a)} \times 2 \text{ (choices for c)} = 6$
The total number of pairs (a,c) for which a+c is even is $8 + 6 = 14$.
For each of these 14 pairs, there are 3 possible choices for the exponent ‘d’ (0, 1, or 2).
Therefore, the total count of divisors in the form (3r+1) is $14 \times 3 = 42$.

Ans: A

Explanation:Let the 4-digit number be represented as $abcd$, where $a$, $b$, $c$, and $d$ are the digits in the thousands, hundreds, tens, and units places, respectively.
We are given the following information:
1. The number is a 4-digit number greater than 1000, which means $a$ can range from 1 to 9, and $b, c, d$ can range from 0 to 9.
2. Sum of the digits in the thousands, hundreds, and tens places is 15: $a + b + c = 15$
3. Sum of the digits in the hundreds, tens, and units places is 16: $b + c + d = 16$
4. The digit in the tens place is 6 more than the digit in the units place: $c = d + 6$

From equation (4), we can determine the possible values for $c$ and $d$. Since $c$ and $d$ are digits from 0 to 9:
If $d = 0$, then $c = 0 + 6 = 6$. This is valid.
If $d = 1$, then $c = 1 + 6 = 7$. This is valid.
If $d = 2$, then $c = 2 + 6 = 8$. This is valid.
If $d = 3$, then $c = 3 + 6 = 9$. This is valid.
If $d > 3$, then $c > 9$, which is not possible for a digit.
So, the possible pairs of $(c, d)$ are $(6, 0), (7, 1), (8, 2), (9, 3)$.

Now, let’s use equation (3) to find the value of $b$ for each pair of $(c, d)$: $b = 16 – c – d$.
Case 1: $(c, d) = (6, 0)$. Then $b = 16 – 6 – 0 = 10$. This is not possible as $b$ must be a single digit.
Case 2: $(c, d) = (7, 1)$. Then $b = 16 – 7 – 1 = 8$. This is valid.
Case 3: $(c, d) = (8, 2)$. Then $b = 16 – 8 – 2 = 6$. This is valid.
Case 4: $(c, d) = (9, 3)$. Then $b = 16 – 9 – 3 = 4$. This is valid.
So, the possible sets of $(b, c, d)$ are $(8, 7, 1), (6, 8, 2), (4, 9, 3)$.

Now, let’s use equation (2) to find the value of $a$ for each set of $(b, c)$: $a = 15 – b – c$. Remember that $a$ must be between 1 and 9.
Case 2a: $(b, c, d) = (8, 7, 1)$. Then $a = 15 – 8 – 7 = 0$. This is not possible as the number is a 4-digit number greater than 1000, so $a$ cannot be 0.
Case 3a: $(b, c, d) = (6, 8, 2)$. Then $a = 15 – 6 – 8 = 1$. This is valid. The number is 1682.
Case 4a: $(b, c, d) = (4, 9, 3)$. Then $a = 15 – 4 – 9 = 2$. This is valid. The number is 2493.

So, the only possible 4-digit numbers that satisfy all the conditions are 1682 and 2493.

The largest possible value of the number is 2493.
The smallest possible value of the number is 1682.

The difference between the largest and smallest possible value of the number is $2493 – 1682 = 811$.

Correct_Option: A

Ans: B

The expression $10^{50} + 10^{25} – 123$ can be rewritten as $10^{50} + (10^{25}-123)$.

Let’s examine the pattern of subtracting 123 from powers of 10:
$10^3 – 123 = 1000 – 123 = 877$
$10^4 – 123 = 10000 – 123 = 9877$
$10^5 – 123 = 100000 – 123 = 99877$
$10^6 – 123 = 1000000 – 123 = 999877$

For an expression of the form $10^n – 123$, which has $n$ digits in total:
The digit $7$ appears $2$ times.
The digit $8$ appears $1$ time.
The digit $9$ appears $(n-3)$ times.

Now, consider the addition to $10^{50}$. The addition of $10^{50}$ to the result of $(10^{25}-123)$ will effectively shift the digits of $(10^{25}-123)$ and add a $1$ as the leftmost digit of the final sum. The digits originating from $(10^{25}-123)$ that are now part of the larger number will contribute their values, but any new digits introduced by the addition of $10^{50}$ to the left of these will be zeros.

Therefore, the sum of the digits of the final result is calculated as follows:
The contribution from the digits of $(10^{25}-123)$ is:
Digits $7$: $7 \times 2 = 14$
Digit $8$: $1 \times 8 = 8$
Digits $9$: $(25-3) \times 9 = 22 \times 9 = 198$
The contribution from the new leading digits due to the addition of $10^{50}$:
The leftmost digit will be $1$.
All other newly introduced digits will be $0$. Their sum contribution is $25 \times 0 = 0$.

Total sum of digits = $14 + 8 + 198 + 0 + 1 = 221$

The correct option is B.

Ans: B

Explanation:To find the remainder when $10^{100}$ is divided by 7, we can use the concept of modular arithmetic and cycles of remainders.

First, let’s find the remainders of powers of 10 when divided by 7:
$10^1 \div 7 = 10 = 1 \times 7 + 3$. The remainder is 3.
$10^2 \div 7 = 100$. $100 = 14 \times 7 + 2$. The remainder is 2.
$10^3 \div 7 = 1000$. $1000 = 142 \times 7 + 6$. The remainder is 6.
Alternatively, $10^3 \equiv 10^2 \times 10 \equiv 2 \times 3 \equiv 6 \pmod{7}$.
$10^4 \div 7$. $10^4 \equiv 10^3 \times 10 \equiv 6 \times 3 \equiv 18 \equiv 2 \times 7 + 4 \pmod{7}$. The remainder is 4.
$10^5 \div 7$. $10^5 \equiv 10^4 \times 10 \equiv 4 \times 3 \equiv 12 \equiv 1 \times 7 + 5 \pmod{7}$. The remainder is 5.
$10^6 \div 7$. $10^6 \equiv 10^5 \times 10 \equiv 5 \times 3 \equiv 15 \equiv 2 \times 7 + 1 \pmod{7}$. The remainder is 1.

We notice a cycle of remainders: 3, 2, 6, 4, 5, 1. The length of this cycle is 6.

To find the remainder of $10^{100}$ when divided by 7, we need to find the position in this cycle corresponding to the exponent 100. We do this by finding the remainder of the exponent 100 when divided by the length of the cycle, which is 6.

$100 \div 6$.
$100 = 16 \times 6 + 4$.
The remainder is 4.

This means that the remainder of $10^{100}$ when divided by 7 will be the same as the 4th remainder in our cycle.

The cycle is (remainder of $10^1$, remainder of $10^2$, remainder of $10^3$, remainder of $10^4$, remainder of $10^5$, remainder of $10^6$):
(3, 2, 6, 4, 5, 1)

Since the remainder of 100 divided by 6 is 4, we look at the 4th element in the cycle, which is 4.

Alternatively, we can use Fermat’s Little Theorem, which states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$.
Here, $a=10$ and $p=7$. Since 7 is prime and 10 is not divisible by 7, we have $10^{7-1} \equiv 10^6 \equiv 1 \pmod{7}$.

Now we want to find the remainder of $10^{100}$ when divided by 7. We can write 100 in terms of powers of 6:
$100 = 16 \times 6 + 4$.
So, $10^{100} = 10^{16 \times 6 + 4} = (10^6)^{16} \times 10^4$.

Using the property of modular arithmetic, if $a \equiv b \pmod{m}$, then $a^k \equiv b^k \pmod{m}$.
Since $10^6 \equiv 1 \pmod{7}$, we have $(10^6)^{16} \equiv 1^{16} \equiv 1 \pmod{7}$.

Therefore, $10^{100} \equiv (10^6)^{16} \times 10^4 \equiv 1 \times 10^4 \pmod{7}$.

Now we need to find the remainder of $10^4$ when divided by 7.
$10^4 \equiv 10 \times 10 \times 10 \times 10 \pmod{7}$
$10 \equiv 3 \pmod{7}$
$10^2 \equiv 3^2 \equiv 9 \equiv 2 \pmod{7}$
$10^3 \equiv 10^2 \times 10 \equiv 2 \times 3 \equiv 6 \pmod{7}$
$10^4 \equiv 10^3 \times 10 \equiv 6 \times 3 \equiv 18 \equiv 4 \pmod{7}$.

So, $10^{100} \equiv 4 \pmod{7}$.

The remainder is 4.

Correct_Option:B

Ans: D

Explanation:Given the equation $m^n = 2^{25} \times 3^{40}$, where $m$ and $n$ are natural numbers and $n > 1$.
We need to express the right-hand side in the form of a power. To do this, we can use the property of exponents $(a^x)^y = a^{xy}$.
We can rewrite the exponents of the prime factors 2 and 3 so that they are divisible by $n$.
Let’s find the greatest common divisor (GCD) of the exponents 25 and 40.
GCD(25, 40) = 5.
This suggests that $n$ could be a factor of 5. Since $n$ is a natural number and $n > 1$, $n$ can be 5.
If $n=5$, we can rewrite the equation as:
$m^5 = 2^{25} \times 3^{40}$
$m^5 = (2^{25/5})^5 \times (3^{40/5})^5$
$m^5 = (2^5)^5 \times (3^8)^5$
Using the property $(a \times b)^x = a^x \times b^x$, we can combine the terms:
$m^5 = (2^5 \times 3^8)^5$
Therefore, $m = 2^5 \times 3^8$.

Now we calculate the values of $m$ and $n$:
$n = 5$
$m = 2^5 \times 3^8 = 32 \times (3^4)^2 = 32 \times (81)^2 = 32 \times 6561$
$m = 32 \times 6561 = 209952$

We need to find $m – n$.
$m – n = 209952 – 5 = 209947$.

Let’s consider if there are other possible values for $n$. $n$ must be a common divisor of 25 and 40, and $n>1$. The common divisors of 25 and 40 are 1 and 5. Since $n>1$, the only possible value for $n$ is 5.

If we tried to express $m^n = 2^{25} \times 3^{40}$ in other ways, say if $n$ was not a common divisor of 25 and 40, we would have fractional exponents for $m$, which is not allowed for natural numbers. For example, if $n=25$, $m = (2^{25} \times 3^{40})^{1/25} = 2 \times 3^{40/25} = 2 \times 3^{8/5}$, which is not a natural number.
Similarly, if $n=40$, $m = (2^{25} \times 3^{40})^{1/40} = 2^{25/40} \times 3 = 2^{5/8} \times 3$, which is not a natural number.
Thus, $n$ must be a common divisor of the exponents, and the largest possible $n$ that allows $m$ to be a natural number is the GCD of the exponents if the exponents are written in their lowest terms relative to $n$.

In this case, $n$ must be a common divisor of 25 and 40. The common divisors are 1 and 5. Since $n > 1$, $n$ must be 5.
$m^5 = 2^{25} \times 3^{40}$
$m = (2^{25} \times 3^{40})^{1/5} = 2^{25/5} \times 3^{40/5} = 2^5 \times 3^8$
$m = 32 \times 6561 = 209952$
$n = 5$
$m – n = 209952 – 5 = 209947$.

Correct_Option: D

Ans: A

Explanation:We want to find the remainder when $3^{333}$ is divided by 11. This is a problem of modular arithmetic. We can use Fermat’s Little Theorem, which states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$.
In this case, $p=11$ (which is prime) and $a=3$ (which is not divisible by 11).
According to Fermat’s Little Theorem, $3^{11-1} \equiv 3^{10} \equiv 1 \pmod{11}$.

Now we need to find the remainder of $3^{333}$ when divided by 11. We can write the exponent 333 in terms of multiples of 10.
$333 = 10 \times 33 + 3$.

So, we can write $3^{333}$ as $3^{10 \times 33 + 3}$.
Using the properties of exponents, this is $(3^{10})^{33} \times 3^3$.

Now we can apply the modular arithmetic:
$3^{333} \equiv (3^{10})^{33} \times 3^3 \pmod{11}$.
Since $3^{10} \equiv 1 \pmod{11}$, we substitute this into the expression:
$3^{333} \equiv (1)^{33} \times 3^3 \pmod{11}$.
$3^{333} \equiv 1 \times 3^3 \pmod{11}$.
$3^{333} \equiv 3^3 \pmod{11}$.

Now, we calculate $3^3$:
$3^3 = 3 \times 3 \times 3 = 9 \times 3 = 27$.

So, we need to find the remainder of 27 when divided by 11.
$27 = 11 \times 2 + 5$.
The remainder is 5.

Therefore, $3^{333} \equiv 5 \pmod{11}$.

The remainder is 5.

Let’s check the options:
Option A: 5
Option B: 10
Option C: 1
Option D: 6

The calculated remainder is 5, which corresponds to Option A.

Correct_Option:A

Ans: C

Explanation:We need to find the remainder when $10^{68}$ is divided by 13. We can use modular arithmetic.
We observe the powers of 10 modulo 13:
$10^1 \equiv 10 \pmod{13}$
$10^2 \equiv 100 \equiv 13 \times 7 + 9 \equiv 9 \pmod{13}$
$10^3 \equiv 10 \times 9 \equiv 90 \equiv 13 \times 6 + 12 \equiv 12 \equiv -1 \pmod{13}$
$10^4 \equiv 10 \times 12 \equiv 120 \equiv 13 \times 9 + 3 \equiv 3 \pmod{13}$
$10^5 \equiv 10 \times 3 \equiv 30 \equiv 13 \times 2 + 4 \equiv 4 \pmod{13}$
$10^6 \equiv 10 \times 4 \equiv 40 \equiv 13 \times 3 + 1 \equiv 1 \pmod{13}$

Since $10^6 \equiv 1 \pmod{13}$, the remainders of powers of 10 when divided by 13 repeat in a cycle of length 6.
We need to find the remainder of $10^{68}$ when divided by 13. We can use the exponent 68 and divide it by the length of the cycle, which is 6.
$68 \div 6$
$68 = 6 \times 11 + 2$
So, $68 \equiv 2 \pmod{6}$.
Therefore, $10^{68} \equiv 10^{6 \times 11 + 2} \equiv (10^6)^{11} \times 10^2 \pmod{13}$.
Since $10^6 \equiv 1 \pmod{13}$, we have:
$10^{68} \equiv 1^{11} \times 10^2 \pmod{13}$
$10^{68} \equiv 1 \times 10^2 \pmod{13}$
$10^{68} \equiv 100 \pmod{13}$
As calculated before, $100 \equiv 9 \pmod{13}$.
Thus, the remainder when $10^{68}$ is divided by 13 is 9.

Alternatively, we can use Fermat’s Little Theorem, which states that if $p$ is a prime number, then for any integer $a$ not divisible by $p$, we have $a^{p-1} \equiv 1 \pmod{p}$.
In this case, $p = 13$ (which is prime) and $a = 10$ (which is not divisible by 13).
So, $10^{13-1} \equiv 10^{12} \equiv 1 \pmod{13}$.
Now we need to find the remainder of $10^{68}$ when divided by 13.
We divide the exponent 68 by 12:
$68 \div 12$
$68 = 12 \times 5 + 8$
So, $68 \equiv 8 \pmod{12}$.
Therefore, $10^{68} \equiv 10^{12 \times 5 + 8} \equiv (10^{12})^5 \times 10^8 \pmod{13}$.
Since $10^{12} \equiv 1 \pmod{13}$:
$10^{68} \equiv 1^5 \times 10^8 \pmod{13}$
$10^{68} \equiv 10^8 \pmod{13}$.
Now we need to find $10^8 \pmod{13}$.
From our earlier calculations:
$10^6 \equiv 1 \pmod{13}$
$10^8 \equiv 10^6 \times 10^2 \pmod{13}$
$10^8 \equiv 1 \times 10^2 \pmod{13}$
$10^8 \equiv 100 \pmod{13}$
$100 \equiv 9 \pmod{13}$.
Thus, the remainder is 9.

Correct_Option:C

Ans: B

Explanation:Let’s first find the prime factorization of 168 and 1134.
$168 = 2 \times 84 = 2 \times 2 \times 42 = 2 \times 2 \times 2 \times 21 = 2^3 \times 3 \times 7$
$1134 = 2 \times 567 = 2 \times 3 \times 189 = 2 \times 3 \times 3 \times 63 = 2 \times 3 \times 3 \times 3 \times 21 = 2 \times 3^4 \times 7$

We are given that 168 is a factor of $1134^n$. This means that the prime factorization of 168 must be contained within the prime factorization of $1134^n$.
$1134^n = (2 \times 3^4 \times 7)^n = 2^n \times 3^{4n} \times 7^n$

For 168 ($2^3 \times 3^1 \times 7^1$) to be a factor of $1134^n$, the exponents of the prime factors in 168 must be less than or equal to the corresponding exponents in $1134^n$.
For prime factor 2: $3 \le n$
For prime factor 3: $1 \le 4n$
For prime factor 7: $1 \le n$

The least positive integer n that satisfies these conditions is the maximum of the lower bounds for n.
$n = \max(3, 1/4, 1) = 3$.
So, the least positive integer n is 3.

Now, we are given that $1134^n$ is a factor of $168^m$.
We have $n=3$, so $1134^3 = (2 \times 3^4 \times 7)^3 = 2^3 \times 3^{12} \times 7^3$.
And $168^m = (2^3 \times 3 \times 7)^m = 2^{3m} \times 3^m \times 7^m$.

For $1134^3$ to be a factor of $168^m$, the exponents of the prime factors in $1134^3$ must be less than or equal to the corresponding exponents in $168^m$.
For prime factor 2: $3 \le 3m \implies 1 \le m$
For prime factor 3: $12 \le m$
For prime factor 7: $3 \le m$

The least positive integer m that satisfies these conditions is the maximum of the lower bounds for m.
$m = \max(1, 12, 3) = 12$.
So, the least positive integer m is 12.

We need to find m + n.
m + n = 12 + 3 = 15.

Correct_Option:B

Ans: C

Explanation:We need to find the number of natural numbers up to 1000 with non-repeating digits.
The numbers can be 1-digit, 2-digit, or 3-digit numbers. The number 1000 itself has repeating digits, so we do not need to consider it.

Case 1: 1-digit numbers
The natural numbers up to 1000 with 1 digit are 1, 2, 3, 4, 5, 6, 7, 8, 9.
All these numbers have non-repeating digits.
Number of 1-digit numbers with non-repeating digits = 9.

Case 2: 2-digit numbers
A 2-digit number has the form AB, where A is the tens digit and B is the units digit.
The tens digit (A) can be any digit from 1 to 9 (since it cannot be 0). So there are 9 choices for A.
The units digit (B) can be any digit from 0 to 9, but it must be different from A. So there are 9 choices for B (10 total digits – 1 digit used for A).
Number of 2-digit numbers with non-repeating digits = 9 (choices for A) * 9 (choices for B) = 81.

Case 3: 3-digit numbers
A 3-digit number has the form ABC, where A is the hundreds digit, B is the tens digit, and C is the units digit.
The hundreds digit (A) can be any digit from 1 to 9 (since it cannot be 0). So there are 9 choices for A.
The tens digit (B) can be any digit from 0 to 9, but it must be different from A. So there are 9 choices for B (10 total digits – 1 digit used for A).
The units digit (C) can be any digit from 0 to 9, but it must be different from both A and B. So there are 8 choices for C (10 total digits – 2 digits used for A and B).
Number of 3-digit numbers with non-repeating digits = 9 (choices for A) * 9 (choices for B) * 8 (choices for C) = 648.

The total number of natural numbers up to 1000 with non-repeating digits is the sum of the numbers from these three cases:
Total = (Number of 1-digit numbers) + (Number of 2-digit numbers) + (Number of 3-digit numbers)
Total = 9 + 81 + 648 = 738.

The number 1000 has repeating digits (0).

Correct_Option:C

Ans: D

It is provided that $a^m \cdot b^n = 144^{145}$, with the conditions $a > 1$ and $b > 1$.
The number $144$ can be expressed as its prime factorization: $144 = 2^4 \times 3^2$.
Therefore, the given equation can be rewritten as:
$a^m \cdot b^n = \left(2^4 \times 3^2\right)^{145} = 2^{4 \times 145} \times 3^{2 \times 145} = 2^{580} \times 3^{290}$
Since $3^{290}$ is a natural number, it can be represented in the form $a^1$, where $a > 1$.
This implies that the minimum possible value for $m$ is 1.
Similarly, the maximum possible value for $n$ can be deduced from the exponent of 2, which is 580.
Consequently, the greatest possible value for the expression $(n-m)$ is calculated as $(580 – 1) = 579$.
The correct choice is D.

Ans: C

Explanation:Let k be a natural number that divides both $m+2n$ and $3m+4n$.
Since k divides $m+2n$, we can write:
$m+2n = ak$ for some integer a. (1)

Since k divides $3m+4n$, we can write:
$3m+4n = bk$ for some integer b. (2)

We want to find a common divisor of some expressions involving m and n. We can use the property that if a number divides two other numbers, it also divides any linear combination of them.

Let’s try to eliminate one of the variables, say m.
Multiply equation (1) by 3:
$3(m+2n) = 3ak$
$3m+6n = 3ak$ (3)

Now subtract equation (2) from equation (3):
$(3m+6n) – (3m+4n) = 3ak – bk$
$3m+6n-3m-4n = (3a-b)k$
$2n = (3a-b)k$
This shows that k divides 2n.

Now let’s try to eliminate n.
Multiply equation (1) by 2:
$2(m+2n) = 2ak$
$2m+4n = 2ak$ (4)

Now subtract equation (4) from equation (2):
$(3m+4n) – (2m+4n) = bk – 2ak$
$3m+4n-2m-4n = (b-2a)k$
$m = (b-2a)k$
This shows that k divides m.

Since k divides m and k divides 2n, k must be a common divisor of m and 2n.

Let’s check the options:
Option A: m and n. k divides m, but not necessarily n.
Option B: 2m and 3n. k divides m, so it divides 2m. k divides 2n, but not necessarily 3n.
Option C: m and 2n. k divides m and k divides 2n. This is what we found.
Option D: 2m and n. k divides m, so it divides 2m. k divides 2n, but not necessarily n.

Therefore, k must be a common divisor of m and 2n.

Correct_Option:C

Ans: 15

As there are two unique divisors apart from 1 and the number itself, the total count of divisors for N is 4. This scenario can materialize in two distinct forms:

Scenario 1: N is the cube of a prime number ($N = p^3$, where $p$ is prime).
Scenario 2: N is the product of two distinct prime numbers ($N = p_1 \times p_2$, where $p_1$ and $p_2$ are prime).

Considering Scenario 1, the numbers that are cubes of a prime and are less than 50 are 8 and 27. This gives us 2 such numbers.

For Scenario 2, we can list the numbers formed by the product of two distinct primes: (2*3), (2*5), (2*7), (2*11), (2*13), (2*17), (2*19), (2*23), (3*5), (3*7), (3*11), (3*13), (5*7). This accounts for 13 numbers.

Therefore, the total count of numbers possessing exactly two distinct divisors is the sum of those from both scenarios: 13 + 2 = 15.

Ans: D

It is stated that precisely 41 numbers, which are powers of two, fall within the range defined by $8^m$ and $8^n$, where $m$ and $n$ are positive integers and $m < n$.

This implies that $2^{3m} < 41 \text{ numbers} < 2^{3n}$.

Given that $m$ is a positive integer, its minimum value is 1. Consequently, $2^{3m}$ is equal to $2^3$. The 41 numbers situated between $2^3$ and $2^{3n}$ must therefore be $2^4, 2^5, 2^6, …, 2^{44}$.

This leads to the conclusion that the lowest possible value for $8^n$ is $2^{45}$. Therefore, the smallest value of $n$ can be determined from $2^{45} = 8^n$, which translates to $2^{3n} = 2^{45}$. Solving for $n$ gives $n = 15$.

The smallest possible value for $m + n$ is therefore $1 + 15 = 16$.

The correct option is D.

Ans: 468

It is established that the count of divisors for these two quantities equals 15. The divisors of 15 are identified as 1, 3, 5, and 15.
The formula for the number of divisors of a number N, represented as $N=a^p\cdot b^q$ (where a and b are prime numbers), is given by $(p+1)\cdot(q+1)$.
To find the smallest possible value of N, we aim for (p+1) and (q+1) to be as close as possible, using the smallest distinct prime numbers for a and b.
Therefore, setting p+1 = 3 (implying p = 2) and q+1 = 5 (implying q = 4), with the prime numbers a and b being 2 and 3 respectively.
Consequently, the minimum value of N is $N=2^4\times\ 3^2\ =144$, and the next smallest value of N is $N=2^2\times\ 3^4\ =324$.
The aggregate of these two values is (144+324) = 468.

Ans: 340

Assume the starting quantity of all fruits is denoted by S.
Let ‘b’ represent the initial quantity of a certain type of fruit, and ‘a’ represent the initial quantity of another type of fruit.
The initial quantity of mangoes is 40% of S, which can be expressed as 2S/5.
The total number of fruits sold is the sum of mangoes sold, apples sold, and bananas sold.
This total is equal to (2S/10) + 96 + (4a/10) = S/2, as given.
Simplifying the equation: S/5 + 96 + 2a/5 = S/2.
Rearranging to solve for S: S = (4a + 960) / 3.
This can also be written as (4a/3) + 320.
For the term (4a/3) to be a whole number, ‘a’ must be divisible by 3.
Additionally, for the term (4a/10) to be a whole number, ‘a’ must be divisible by 5.
The smallest positive integer value for ‘a’ that satisfies both divisibility conditions is 15.
Substituting this value back into the expression for S: S = (4 * 15) / 3 + 320 = 60 / 3 + 320 = 20 + 320 = 340.

Ans: 82

Let A represent the greatest common divisor (GCD) of the expression $3^k+4^k+5^k$ for various integer values of k.

When k = 1, the expression evaluates to $3^1+4^1+5^1 = 3+4+5 = 12$.
When k = 2, the expression evaluates to $3^2+4^2+5^2 = 9+16+25 = 50$.
When k = 3, the expression evaluates to $3^3+4^3+5^3 = 27+64+125 = 216$.

The GCD of these computed values (12, 50, 216) is 2. Thus, A = 2.

Let B be the GCD of the expression $4^k+3\left(4^k\right)+4^{k+2}$ for different integer values of k.
We can simplify this expression:
$4^k+3\left(4^k\right)+4^{k+2} = 4^k(1+3) + 4^{k+2}$
$= 4^k(4) + 4^{k+2}$
$= 4^{k+1} + 4^{k+2}$
$= 4^{k+1}(1+4)$
$= 5 \cdot 4^{k+1}$

The GCD of these values is determined by the smallest possible value of k.
When k = 1, the expression evaluates to $5 \cdot 4^{1+1} = 5 \cdot 4^2 = 5 \cdot 16 = 80$.
Therefore, B = 80.

The sum A + B is $2 + 80 = 82$.

Ans: B

To determine the greatest possible integer ‘n’ such that n factorial (n!) does not exceed 15000:

Calculate the factorial values:
7! = 5040
8! = 40320

Observe that 8! exceeds 15000. This means that any factorial greater than 7! will also be greater than 15000. Consequently, 15000 cannot be divisible by 8! or any subsequent factorial.

Thus, the largest integer ‘n’ for which n! is less than 15000 is 7.

This corresponds to option B.

Ans: 150

When the total student count is divided by 9, 10, 12, or 25, a remainder of 4 is consistently observed. This implies the total number of students can be expressed as the Least Common Multiple (LCM) of 9, 10, 12, and 25, multiplied by an integer ‘k’, plus 4. The LCM(9, 10, 12, 25) is 900, so the number of students is of the form 900k + 4.

We are informed that this total student count (900k + 4) is less than 5000. Additionally, the same count, 900k + 4, yields a specific remainder when divided by 11. This condition is satisfied when k equals 2, resulting in a total of 1804 students (900 * 2 + 4 = 1804).

Therefore, the number of groups of 12 students would be 1800 divided by 12, which equals 150.

Ans: 70

Explanation:Let the three-digit number be represented as $100a + 10b + c$, where $a$, $b$, and $c$ are digits, and $a \neq 0$.
The number is greater than 100, which is already satisfied since $a \neq 0$ for a three-digit number.
When the digits are arranged in the reverse order, the new number is $100c + 10b + a$.

According to the problem statement, the reversed number is 198 greater than the original number.
So, we have the equation:
$100c + 10b + a = (100a + 10b + c) + 198$

Now, let’s simplify the equation:
$100c + 10b + a – 100a – 10b – c = 198$
Combine like terms:
$(100c – c) + (10b – 10b) + (a – 100a) = 198$
$99c + 0 – 99a = 198$
$99(c – a) = 198$

Divide both sides by 99:
$c – a = \frac{198}{99}$
$c – a = 2$

This equation tells us that the difference between the units digit ($c$) and the hundreds digit ($a$) must be 2.
The digits $a$, $b$, and $c$ must be integers from 0 to 9.
Also, $a$ cannot be 0 because it is the hundreds digit of a three-digit number.

Let’s find possible pairs of $(a, c)$ that satisfy $c – a = 2$, with $a \in \{1, 2, …, 9\}$ and $c \in \{0, 1, …, 9\}$.

If $a = 1$, then $c = 1 + 2 = 3$. Possible number: $1b3$.
If $a = 2$, then $c = 2 + 2 = 4$. Possible number: $2b4$.
If $a = 3$, then $c = 3 + 2 = 5$. Possible number: $3b5$.
If $a = 4$, then $c = 4 + 2 = 6$. Possible number: $4b6$.
If $a = 5$, then $c = 5 + 2 = 7$. Possible number: $5b7$.
If $a = 6$, then $c = 6 + 2 = 8$. Possible number: $6b8$.
If $a = 7$, then $c = 7 + 2 = 9$. Possible number: $7b9$.

If $a = 8$, then $c = 8 + 2 = 10$, which is not a valid digit.
If $a = 9$, then $c = 9 + 2 = 11$, which is not a valid digit.

So, the possible pairs for $(a, c)$ are (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8), and (7, 9).
There are 7 such pairs.

The digit $b$ can be any digit from 0 to 9, as it does not affect the difference between the number and its reverse.
For each valid pair of $(a, c)$, there are 10 possible values for $b$ (0, 1, 2, 3, 4, 5, 6, 7, 8, 9).

Since there are 7 possible pairs for $(a, c)$ and 10 possible values for $b$ for each pair, the total number of such three-digit numbers is $7 \times 10 = 70$.

Final_Answer:70

Ans: 4195

Let’s break down a four-digit numerical puzzle.
Imagine a number represented by its digits: ‘a’ in the thousands place, ‘b’ in the hundreds place, ‘c’ in the tens place, and ‘d’ in the units place. So, the number is conceptually ‘abcd’.

We are given three conditions:
1. The sum of the first three digits is 14: a + b + c = 14.
2. The sum of the last three digits is 15: b + c + d = 15.
3. The tens digit is 4 more than the units digit: c = d + 4.

Our goal is to find the largest possible four-digit number that fits these rules. To maximize the number ‘abcd’, we should aim for the largest possible value for ‘a’, the thousands digit. Remember that ‘a’, ‘b’, ‘c’, and ‘d’ are all single digits, meaning they can be any integer from 0 to 9.

Let’s start manipulating the equations:
Substitute the third condition (c = d + 4) into the second condition (b + c + d = 15):
b + (d + 4) + d = 15
This simplifies to: b + 2d = 11. (Let’s call this equation 4)

Now, let’s find a relationship between ‘a’ and ‘d’. Subtract the first equation (a + b + c = 14) from the second equation (b + c + d = 15):
(b + c + d) – (a + b + c) = 15 – 14
This results in: d = a + 1. (Let’s call this equation 5)

To find the maximum possible number, we need to maximize ‘a’. Since ‘c’ is a single digit, its maximum possible value is 9.
Using the condition c = d + 4, if c = 9, then d must be 5 (because 9 = 5 + 4).
Now that we know d = 5, we can use equation 5 (d = a + 1) to find ‘a’. If d = 5, then a = 4 (because 5 = 4 + 1).
So, the maximum possible value for ‘a’ is 4 when ‘c’ is its maximum (9) and ‘d’ is consequently 5.

Now we need to find ‘b’. We can use equation 4 (b + 2d = 11). With d = 5, the equation becomes:
b + 2(5) = 11
b + 10 = 11
b = 1.

Therefore, the four digits are a=4, b=1, c=9, and d=5. The largest four-digit number that satisfies all the given conditions is 4195.

Ans: 21

Explanation:Let the 3-digit number be represented by its digits $a$, $b$, and $c$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit.
The digits $a, b, c$ must be integers such that $1 \le a \le 9$ and $0 \le b, c \le 9$.
We are given the condition that the product of the digits is more than 2 but less than 7.
So, $2 < a \cdot b \cdot c < 7$.
This means the product $a \cdot b \cdot c$ can be 3, 4, 5, or 6.

Case 1: $a \cdot b \cdot c = 3$.
Since 3 is a prime number, the possible sets of digits are {1, 1, 3}.
For the digits to form a 3-digit number, the hundreds digit ($a$) cannot be 0.
The possible permutations of {1, 1, 3} are:
113, 131, 311.
There are 3 such numbers.

Case 2: $a \cdot b \cdot c = 4$.
The possible sets of digits whose product is 4 are:
– {1, 1, 4}
Permutations: 114, 141, 411 (3 numbers)
– {1, 2, 2}
Permutations: 122, 212, 221 (3 numbers)
Total numbers for product 4 = 3 + 3 = 6.

Case 3: $a \cdot b \cdot c = 5$.
Since 5 is a prime number, the possible sets of digits are {1, 1, 5}.
The possible permutations of {1, 1, 5} are:
115, 151, 511.
There are 3 such numbers.

Case 4: $a \cdot b \cdot c = 6$.
The possible sets of digits whose product is 6 are:
– {1, 1, 6}
Permutations: 116, 161, 611 (3 numbers)
– {1, 2, 3}
Permutations: 123, 132, 213, 231, 312, 321 (6 numbers)
Total numbers for product 6 = 3 + 6 = 9.

Total number of 3-digit numbers = (Numbers for product 3) + (Numbers for product 4) + (Numbers for product 5) + (Numbers for product 6)
Total = 3 + 6 + 3 + 9 = 21.

We need to ensure that none of the digits are 0.
In Case 1 ($a \cdot b \cdot c = 3$), the digits are {1, 1, 3}. No zeros.
In Case 2 ($a \cdot b \cdot c = 4$), the digits are {1, 1, 4} and {1, 2, 2}. No zeros.
In Case 3 ($a \cdot b \cdot c = 5$), the digits are {1, 1, 5}. No zeros.
In Case 4 ($a \cdot b \cdot c = 6$), the digits are {1, 1, 6} and {1, 2, 3}. No zeros.
Since none of the possible combinations of digits for products 3, 4, 5, and 6 include 0, all the permutations will result in valid 3-digit numbers where the hundreds digit is not 0.

Therefore, the total count of such 3-digit numbers is 21.

Final_Answer:21

Ans: D

Explanation:We are given that a, b, and c are positive integers.
We have the equations:
1. ab = 432
2. bc = 96
3. c < 9

From equation (2), we can express b in terms of c:
b = 96/c

Since b is a positive integer, c must be a divisor of 96.
Also, we are given that c < 9 and c is a positive integer. So the possible values for c are 1, 2, 3, 4, 5, 6, 7, 8.
We need to find the divisors of 96 among these values.
The divisors of 96 are 1, 2, 3, 4, 6, 8.
So the possible values for c are 1, 2, 3, 4, 6, 8.

Now, we can find the corresponding values of b for each possible value of c using b = 96/c:
– If c = 1, b = 96/1 = 96
– If c = 2, b = 96/2 = 48
– If c = 3, b = 96/3 = 32
– If c = 4, b = 96/4 = 24
– If c = 6, b = 96/6 = 16
– If c = 8, b = 96/8 = 12

Next, we use equation (1), ab = 432, to find the corresponding values of a using a = 432/b. Since a is a positive integer, b must be a divisor of 432. Let’s check if our calculated values of b are divisors of 432.
– If b = 96, a = 432/96. 432 = 4 * 96 + 48. So 432/96 is not an integer. This case is not possible.
– If b = 48, a = 432/48. 432 = 9 * 48. So a = 9. (c=2, b=48, a=9)
– If b = 32, a = 432/32. 432 = 13 * 32 + 16. So 432/32 is not an integer. This case is not possible.
– If b = 24, a = 432/24. 432 = 18 * 24. So a = 18. (c=4, b=24, a=18)
– If b = 16, a = 432/16. 432 = 27 * 16. So a = 27. (c=6, b=16, a=27)
– If b = 12, a = 432/12. 432 = 36 * 12. So a = 36. (c=8, b=12, a=36)

Now we calculate the sum a + b + c for the valid combinations:
– For c = 2, b = 48, a = 9: a + b + c = 9 + 48 + 2 = 59
– For c = 4, b = 24, a = 18: a + b + c = 18 + 24 + 4 = 46
– For c = 6, b = 16, a = 27: a + b + c = 27 + 16 + 6 = 49
– For c = 8, b = 12, a = 36: a + b + c = 36 + 12 + 8 = 56

The possible values for a + b + c are 59, 46, 49, and 56.
The smallest possible value among these is 46.

Correct_Option:D

Ans: 252

Explanation:The problem asks for the number of integers in the set {100, 101, 102, …, 999} that have at least one digit repeated.
This set consists of all three-digit integers. The total number of integers in this set is 999 – 100 + 1 = 900.

It is easier to find the number of integers with no repeated digits and then subtract this from the total number of integers.

Let’s find the number of three-digit integers with no repeated digits.
A three-digit integer can be represented as $abc$, where $a$ is the hundreds digit, $b$ is the tens digit, and $c$ is the units digit.

For the hundreds digit ($a$), it can be any digit from 1 to 9 (since it’s a three-digit number, the first digit cannot be 0). So there are 9 choices for $a$.

For the tens digit ($b$), it can be any digit from 0 to 9, except for the digit already chosen for $a$. So there are 10 – 1 = 9 choices for $b$.

For the units digit ($c$), it can be any digit from 0 to 9, except for the digits already chosen for $a$ and $b$. Since $a$ and $b$ are distinct, we have used 2 distinct digits. So there are 10 – 2 = 8 choices for $c$.

The total number of three-digit integers with no repeated digits is the product of the number of choices for each digit:
Number of integers with no repeated digits = (Choices for $a$) * (Choices for $b$) * (Choices for $c$)
= 9 * 9 * 8
= 81 * 8
= 648

Now, to find the number of integers with at least one digit repeated, we subtract the number of integers with no repeated digits from the total number of integers in the set:
Number of integers with at least one digit repeated = (Total number of integers) – (Number of integers with no repeated digits)
= 900 – 648
= 252

Final_Answer:252

Ans: B

To determine the count of integers from 1 to 120 that are not divisible by 2, 5, or 7, we can follow these steps:

1. **Identify multiples of 2:** The quantity of numbers divisible by 2 within the range of 1 to 120 is 60.

2. **Identify multiples of 5, excluding multiples of 2:** The count of numbers divisible by 5 but not by 2 within the range of 1 to 120 is 12.

3. **Identify multiples of 7, excluding multiples of 2 and 5:** The quantity of numbers divisible by 7 but not by 2 or 5 within the range of 1 to 120 is 7.

4. **Calculate the total:** Subtract the counts from the previous steps from the total number of integers in the range. Therefore, the number of integers from 1 to 120 that are divisible by none of 2, 5, and 7 is 120 – 60 – 12 – 7 = 41.

Ans: A

Explanation:We are looking for the number of pairs of positive integers $(a, b)$ such that $a \leq b$ and $ab = 4^{2017}$.
First, let’s express $4^{2017}$ in terms of its prime factorization.
$4^{2017} = (2^2)^{2017} = 2^{2 \times 2017} = 2^{4034}$.

We need to find pairs of positive integers $(a, b)$ such that $ab = 2^{4034}$.
Let $a = 2^x$ and $b = 2^y$, where $x$ and $y$ are non-negative integers.
Then $ab = 2^x \cdot 2^y = 2^{x+y}$.
So, we must have $x+y = 4034$.

Since $a$ and $b$ are positive integers, $x$ and $y$ must be non-negative integers.
The possible values for $x$ are $0, 1, 2, \ldots, 4034$.
For each value of $x$, $y$ is uniquely determined as $y = 4034 – x$.

The condition $a \leq b$ translates to $2^x \leq 2^y$, which means $x \leq y$.
Substituting $y = 4034 – x$, we get $x \leq 4034 – x$.
Adding $x$ to both sides, we have $2x \leq 4034$.
Dividing by 2, we get $x \leq 2017$.

So, the possible values for $x$ are integers such that $0 \leq x \leq 2017$.
The possible integer values for $x$ are $0, 1, 2, \ldots, 2017$.
The number of such values is $2017 – 0 + 1 = 2018$.

Each value of $x$ in this range corresponds to a unique pair $(a, b)$ satisfying the given conditions:
If $x = 0$, $a = 2^0 = 1$, $y = 4034$, $b = 2^{4034}$. Pair: $(1, 2^{4034})$. $1 \leq 2^{4034}$.
If $x = 1$, $a = 2^1 = 2$, $y = 4033$, $b = 2^{4033}$. Pair: $(2, 2^{4033})$. $2 \leq 2^{4033}$.
…
If $x = 2017$, $a = 2^{2017}$, $y = 4034 – 2017 = 2017$, $b = 2^{2017}$. Pair: $(2^{2017}, 2^{2017})$. $2^{2017} \leq 2^{2017}$.

If $x > 2017$, then $x > 4034 – x$, which means $a > b$, violating the condition $a \leq b$.

The number of possible values for $x$ is 2018. Therefore, there are 2018 such pairs $(a, b)$.

The final answer is $\boxed{2018}$.

Correct_Option:A

Ans: D

Explanation:Let the given expression be $k$.
$k = \frac{n^2 + 7n + 12}{n^2 – n – 12}$
First, we factorize the numerator and the denominator.
Numerator: $n^2 + 7n + 12 = (n+3)(n+4)$
Denominator: $n^2 – n – 12 = (n-4)(n+3)$

So, the expression becomes:
$k = \frac{(n+3)(n+4)}{(n-4)(n+3)}$
For this expression to be defined, the denominator cannot be zero, which means $(n-4)(n+3) \neq 0$. Thus, $n \neq 4$ and $n \neq -3$.

Since we are looking for a positive integer $n$, $n+3$ will always be positive. We can cancel out the $(n+3)$ term from the numerator and the denominator, provided $n \neq -3$.
$k = \frac{n+4}{n-4}$
We can rewrite this expression by using polynomial division or algebraic manipulation:
$k = \frac{n-4+8}{n-4} = \frac{n-4}{n-4} + \frac{8}{n-4} = 1 + \frac{8}{n-4}$
For $k$ to be a positive integer, $1 + \frac{8}{n-4}$ must be a positive integer. This means $\frac{8}{n-4}$ must be an integer such that $1 + \frac{8}{n-4} > 0$.

For $\frac{8}{n-4}$ to be an integer, $(n-4)$ must be a divisor of 8. The divisors of 8 are $\{-8, -4, -2, -1, 1, 2, 4, 8\}$.

We have the following possible values for $n-4$:
1. $n-4 = -8 \implies n = -4$. This is not a positive integer.
2. $n-4 = -4 \implies n = 0$. This is not a positive integer.
3. $n-4 = -2 \implies n = 2$. In this case, $k = 1 + \frac{8}{-2} = 1 – 4 = -3$. This is not a positive integer.
4. $n-4 = -1 \implies n = 3$. In this case, $k = 1 + \frac{8}{-1} = 1 – 8 = -7$. This is not a positive integer.
5. $n-4 = 1 \implies n = 5$. In this case, $k = 1 + \frac{8}{1} = 1 + 8 = 9$. This is a positive integer.
6. $n-4 = 2 \implies n = 6$. In this case, $k = 1 + \frac{8}{2} = 1 + 4 = 5$. This is a positive integer.
7. $n-4 = 4 \implies n = 8$. In this case, $k = 1 + \frac{8}{4} = 1 + 2 = 3$. This is a positive integer.
8. $n-4 = 8 \implies n = 12$. In this case, $k = 1 + \frac{8}{8} = 1 + 1 = 2$. This is a positive integer.

We are looking for the largest positive integer $n$. From the above list of positive integer values for $n$ that result in a positive integer $k$, we have $n \in \{5, 6, 8, 12\}$.
The largest among these values is 12.

We also need to ensure that the original denominator is not zero.
For $n=12$, $n^2 – n – 12 = 12^2 – 12 – 12 = 144 – 24 = 120 \neq 0$.
The original expression is $\frac{12^2 + 7(12) + 12}{12^2 – 12 – 12} = \frac{144 + 84 + 12}{144 – 12 – 12} = \frac{240}{120} = 2$, which is a positive integer.

Let’s check the options provided:
Option A: n=6. k = 1 + 8/(6-4) = 1 + 8/2 = 1 + 4 = 5. Positive integer.
Option B: n=16. k = 1 + 8/(16-4) = 1 + 8/12 = 1 + 2/3. Not an integer.
Option C: n=8. k = 1 + 8/(8-4) = 1 + 8/4 = 1 + 2 = 3. Positive integer.
Option D: n=12. k = 1 + 8/(12-4) = 1 + 8/8 = 1 + 1 = 2. Positive integer.

The positive integer values of n for which k is a positive integer are 5, 6, 8, 12.
The largest of these is 12.

Correct_Option:D

Ans: 44

$2^4 \times 3^5 \times 10^4$
$= 2^4 \times 3^5 \times (2 \times 5)^4$
$= 2^4 \times 3^5 \times 2^4 \times 5^4$
$= 2^{(4+4)} \times 3^5 \times 5^4$
$= 2^8 \times 3^5 \times 5^4$

To form a perfect square, each prime factor must have an even exponent.

Considering the prime factor 2, with an exponent of 8 ($2^8$), the possible even exponents are 0, 2, 4, 6, and 8. This gives us $2^0, 2^2, 2^4, 2^6, 2^8$ – a total of 5 possibilities.

For the prime factor 3, with an exponent of 5 ($3^5$), the possible even exponents are 0, 2, and 4. This gives us $3^0, 3^2, 3^4$ – a total of 3 possibilities.

For the prime factor 5, with an exponent of 4 ($5^4$), the possible even exponents are 0, 2, and 4. This gives us $5^0, 5^2, 5^4$ – a total of 3 possibilities.

The total number of perfect square factors is the product of the number of possibilities for each prime factor: $5 \times 3 \times 3 = 45$.

However, this count includes the factor $2^0 \times 3^0 \times 5^0 = 1$, which is considered a perfect square. If we need to find the number of perfect squares *greater than 1*, we subtract this one case.

Number of perfect squares greater than 1 = $45 – 1 = 44$.

Ans: 7

Let the six-digit number be represented by its digits as $d_1d_2d_3d_4d_5d_6$.
The given conditions can be translated as:
$d_6 = d_1 + d_2 + d_3$
$d_5 = d_1 + d_2$
$d_3 = d_1$
$d_2 = 2 \times d_1$
$d_4 = d_5 + d_6$

Substituting the relationships into the equation for $d_4$:
$d_4 = (d_1 + d_2) + (d_1 + d_2 + d_3)$
$d_4 = d_1 + d_2 + d_1 + d_2 + d_3$
$d_4 = 2d_1 + 2d_2 + d_3$

Now, substitute the relationships for $d_2$ and $d_3$ in terms of $d_1$:
$d_2 = 2d_1$
$d_3 = d_1$

So, $d_4 = 2d_1 + 2(2d_1) + d_1$
$d_4 = 2d_1 + 4d_1 + d_1$
$d_4 = 7d_1$

Since the number is a six-digit number, the first digit, $d_1$, cannot be zero.
Also, $d_4$ is a single digit. If $d_1$ were 2 or greater, $d_4$ would be $7 \times 2 = 14$ or greater, which is a two-digit number, making it impossible for $d_4$ to be a single digit.
Therefore, $d_1$ must be 1.
This implies $d_4 = 7 \times 1 = 7$.

Ans: 6

Explanation:Let the two-digit number be represented as $10a + b$, where $a$ is the digit in the tens place and $b$ is the digit in the units place.
The digits $a$ and $b$ are integers.
Since it is a two-digit number, $a$ must be non-zero, so $a \in \{1, 2, …, 9\}$.
The units digit is non-zero, so $b \in \{1, 2, …, 9\}$.

The number formed by interchanging the positions of its digits is $10b + a$.

The problem states that the original number is more than thrice the number formed by interchanging its digits.
So, we have the inequality:
$10a + b > 3(10b + a)$

Now, let’s solve this inequality:
$10a + b > 30b + 3a$
Subtract $3a$ from both sides:
$7a + b > 30b$
Subtract $b$ from both sides:
$7a > 29b$

We need to find the number of pairs of integers $(a, b)$ such that $a \in \{1, 2, …, 9\}$, $b \in \{1, 2, …, 9\}$, and $7a > 29b$.

Let’s test values for $b$ from 1 to 9 and see the possible values of $a$:

Case 1: $b = 1$
$7a > 29(1)$
$7a > 29$
$a > \frac{29}{7}$
$a > 4.14…$
Since $a$ must be an integer and $a \in \{1, 2, …, 9\}$, the possible values for $a$ are $\{5, 6, 7, 8, 9\}$.
This gives 5 numbers: 51, 61, 71, 81, 91.

Case 2: $b = 2$
$7a > 29(2)$
$7a > 58$
$a > \frac{58}{7}$
$a > 8.28…$
Since $a$ must be an integer and $a \in \{1, 2, …, 9\}$, the only possible value for $a$ is $\{9\}$.
This gives 1 number: 92.

Case 3: $b = 3$
$7a > 29(3)$
$7a > 87$
$a > \frac{87}{7}$
$a > 12.42…$
Since the maximum value for $a$ is 9, there are no possible values for $a$ in this case.

For any $b > 2$, the value of $29b$ will be even larger, and thus $a$ will need to be even larger than 12.42…, meaning there will be no solutions for $a$ in the range $\{1, 2, …, 9\}$.

So, the possible pairs $(a, b)$ are:
From Case 1 ($b=1$): (5,1), (6,1), (7,1), (8,1), (9,1)
From Case 2 ($b=2$): (9,2)

The two-digit numbers are:
For $b=1$: 51, 61, 71, 81, 91
For $b=2$: 92

The total number of such two-digit numbers is $5 + 1 = 6$.

Let’s verify for a few numbers.
For 51: Interchanged is 15. 3 * 15 = 45. 51 > 45. (Correct)
For 91: Interchanged is 19. 3 * 19 = 57. 91 > 57. (Correct)
For 92: Interchanged is 29. 3 * 29 = 87. 92 > 87. (Correct)

Final Answer: The total count of such numbers is 6.

The set of numbers are {51, 61, 71, 81, 91, 92}.

Final_Answer: 6

Ans: 10

Explanation:We are given that $N$ and $x$ are positive integers such that $N^N = 2^{160}$.
We need to find the largest possible value of $x$ such that $N^2 + 2^N$ is an integral multiple of $2^x$.

From the equation $N^N = 2^{160}$, we can deduce that $N$ must be a power of 2. Let $N = 2^k$ for some positive integer $k$.
Then $(2^k)^{2^k} = 2^{160}$, which means $2^{k \cdot 2^k} = 2^{160}$.
Equating the exponents, we get $k \cdot 2^k = 160$.

We need to find an integer $k$ that satisfies this equation. Let’s test some values of $k$:
If $k=1$, $1 \cdot 2^1 = 2 \neq 160$.
If $k=2$, $2 \cdot 2^2 = 2 \cdot 4 = 8 \neq 160$.
If $k=3$, $3 \cdot 2^3 = 3 \cdot 8 = 24 \neq 160$.
If $k=4$, $4 \cdot 2^4 = 4 \cdot 16 = 64 \neq 160$.
If $k=5$, $5 \cdot 2^5 = 5 \cdot 32 = 160$.
So, $k=5$ is the solution.

Therefore, $N = 2^k = 2^5 = 32$.

Now we need to find the largest possible $x$ such that $N^2 + 2^N$ is an integral multiple of $2^x$.
Substitute $N=32$ into the expression:
$N^2 + 2^N = 32^2 + 2^{32}$
$32^2 = (2^5)^2 = 2^{10}$
So, $N^2 + 2^N = 2^{10} + 2^{32}$.

We can factor out the lowest power of 2 from this expression:
$2^{10} + 2^{32} = 2^{10} (1 + 2^{32-10}) = 2^{10} (1 + 2^{22})$

We are given that $N^2 + 2^N$ is an integral multiple of $2^x$. This means $2^{10} (1 + 2^{22})$ is divisible by $2^x$.
The term $(1 + 2^{22})$ is an odd number because $2^{22}$ is an even number, and adding 1 makes it odd. Therefore, $(1 + 2^{22})$ is not divisible by 2.
The highest power of 2 that divides $2^{10} (1 + 2^{22})$ is $2^{10}$.

So, $2^{10} (1 + 2^{22}) = m \cdot 2^x$, where $m$ is an integer.
Since $(1 + 2^{22})$ is odd, for the divisibility to hold, $2^x$ must divide $2^{10}$.
The largest power of 2 that divides $2^{10}$ is $2^{10}$.
Thus, the largest possible value of $x$ is 10.

Final_Answer:10

Ans: 3

Explanation:This is a classic magic square problem. In a 3×3 magic square using numbers 1 through 9, the sum of each row, column, and diagonal is constant. This constant sum, often called the magic constant, can be calculated. The sum of numbers from 1 to 9 is (9 * 10) / 2 = 45. Since there are 3 rows (or columns), the magic constant is 45 / 3 = 15.

Let the grid be represented as:
a b c
d e f
g h i

We are given that the numbers 1 through 9 are used exactly once.
We are given:
a = 6
c = 2

The magic sum for each row, column, and diagonal is 15.

From the top row: a + b + c = 15
Substituting the given values: 6 + b + 2 = 15
8 + b = 15
b = 15 – 8
b = 7

Now the top row is:
6 7 2

Let’s consider the columns and diagonals.
The middle element (e) of a 3×3 magic square using numbers 1-9 is always the average of the numbers, which is (1+9)/2 = 5. So, e = 5.

Now we have:
6 7 2
d 5 f
g h i

Let’s use the column sums:
Column 1: a + d + g = 15 => 6 + d + g = 15 => d + g = 9
Column 2: b + e + h = 15 => 7 + 5 + h = 15 => 12 + h = 15 => h = 3
Column 3: c + f + i = 15 => 2 + f + i = 15 => f + i = 13

Let’s use the diagonal sums:
Diagonal 1 (top-left to bottom-right): a + e + i = 15 => 6 + 5 + i = 15 => 11 + i = 15 => i = 4
Diagonal 2 (top-right to bottom-left): c + e + g = 15 => 2 + 5 + g = 15 => 7 + g = 15 => g = 8

Now we have:
a=6, b=7, c=2
d=?, e=5, f=?
g=8, h=3, i=4

We can find ‘d’ and ‘f’ using the remaining equations or by ensuring all numbers are used.
From Column 1: d + g = 9 => d + 8 = 9 => d = 1
From Column 3: f + i = 13 => f + 4 = 13 => f = 9

The complete magic square is:
6 7 2
1 5 9
8 3 4

Let’s verify all sums:
Rows:
6 + 7 + 2 = 15
1 + 5 + 9 = 15
8 + 3 + 4 = 15

Columns:
6 + 1 + 8 = 15
7 + 5 + 3 = 15
2 + 9 + 4 = 15

Diagonals:
6 + 5 + 4 = 15
2 + 5 + 8 = 15

The numbers used are 1, 2, 3, 4, 5, 6, 7, 8, 9, each once.
The question asks for the bottom middle entry, which is ‘h’.
We found h = 3.

Final Answer:3
Final_Answer:3

Ans: D

Explanation:Let the three consecutive positive integers be $n-1$, $n$, and $n+1$.
Their product is given as 15600.
So, $(n-1) \times n \times (n+1) = 15600$.
This can be written as $n(n^2 – 1) = 15600$.
We need to find an integer $n$ such that $n^3$ is slightly greater than 15600.
Let’s estimate the value of $n$.
$20^3 = 8000$
$30^3 = 27000$
So, $n$ is between 20 and 30.
Let’s try values near the cube root of 15600.
$\sqrt[3]{15600} \approx 25$
Let’s check if $n=25$ works.
If $n=25$, the three integers are 24, 25, and 26.
Their product is $24 \times 25 \times 26$.
$24 \times 25 = 600$
$600 \times 26 = 15600$.
So, the three consecutive positive integers are 24, 25, and 26.

We are asked to find the sum of the squares of these integers.
Sum of squares $= 24^2 + 25^2 + 26^2$.
$24^2 = 576$
$25^2 = 625$
$26^2 = 676$
Sum of squares $= 576 + 625 + 676 = 1877$.

Alternatively, we can use the algebraic expression for the sum of squares:
Sum of squares $= (n-1)^2 + n^2 + (n+1)^2$
$= (n^2 – 2n + 1) + n^2 + (n^2 + 2n + 1)$
$= 3n^2 + 2$.
Since $n=25$, the sum of squares $= 3 \times 25^2 + 2$.
$= 3 \times 625 + 2$
$= 1875 + 2$
$= 1877$.

Correct_Option:D