To solve this question, we need to immediately recognise the fact that,&nbsp;$32768=8^5$Substituting this in the above given equation,&nbsp;$\left(\dfrac{1}{8}\right)^k\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{3}}$ = $\left(\dfrac{1}{8}\right)\times\ \left(\dfrac{1}{8}\right)^{5\times\ \dfrac{1}{k}}$Since the bases are equal, we can equate the powers on either side of the equation,&nbsp;$k+\frac{5}{3}=1+\frac{5}{k}$$\frac{\left(3k+5\right)}{3}=\frac{\left(k+5\right)}{k}$$3k^2+5k=3k+15$$3k^2+2k-15=0$Here in the given quadratic equation, the Discriminant is greater than 0,&nbsp;$2^2-\left(4\right)\left(3\right)\left(-15\right)$&gt; 0That means both the roots are real, hence we can simply take the sum of the roots of the quadratic equation in k,&nbsp;Which in a standard quadratic equation of the form&nbsp;$ax^2+bx+c$ is&nbsp;$-\frac{b}{a}$Here, the sum of the real values of k is&nbsp;$-\frac{2}{3}$

When gives n distinct numbers, and asked to find the sum of the possible n distinct numbers that can be formed,&nbsp;We use the formula,&nbsp;$\left(10^{n-1}+10^{n-2}+10^{n-3}+...\right)\left(\left(n-1\right)!\right)\left(Sum\ of\ numbers\right)$Inserting n=4,&nbsp;$\left(1000+100+10+1\right)\left(3!\right)\left(a+b+c+d\right)$$\left(6666\right)\left(a+b+c+d\right)$We are told that this value equals,&nbsp;$153310+n$Since we are told that n is a single digit natural number, the total value cannot be that much greater and 6666 should perfectly divide&nbsp;$153310+n$Upon dividing 153310 by 6666 we get the quotient as 22.99Nearest value being 23, We take 6666x23 giving us the value 153318Hence the value of n=8 and the value of a+b+c+d=23Value of&nbsp;a+b+c+d+n=31.&nbsp;

We must bring the right-hand side in the form so that everything has the same power.&nbsp;25 has factors 1, 5 and 25The only common factor 40 and 25 have is 5 (other than 1 of course, which does not work)So the right-hand side can be rewritten as&nbsp;$\left(2^5\right)^5\times\ \left(3^8\right)^5$$\left(32\times\ 81\times\ 81\right)^5$$\left(209952\right)^5$Giving the value of m - n as 209952 - 5 = 209947Therefore, Option D is the correct answer.&nbsp;

We need to consider single-digit, two-digit and three-digit numbersSingle digit:&nbsp;There are only nine numbers for a single-digit number (not including zero since we are looking for positive integers only)Double digits: The ten's digit can be chosen in 9 ways (not including 0), and the unit digit can be chosen in 9 ways (including 0 but excluding the digit used in the ten's place).&nbsp;Hence, a total of 81 numbers.&nbsp;Three-digit:&nbsp;The hundred's digit can be 1, 2, 3 or 4; hence, there&nbsp;are four options for the hundred's digit; for the ten's digit, there will be 9 options (numbers from 0 to 9 except the one chosen in the hundred's place); for unit's digit there would be 8 options. Hence, a total of 288 numbersGiving the total numbers as 288+81+9 = 378Therefore, 378 is the correct answer.&nbsp;

Prime factorising 1134, we get&nbsp;&nbsp;$1134 = 2 \times 3^4 \times 7$&nbsp;&nbsp;and&nbsp;&nbsp;$168 = 2^3 \times 3 \times 7$$1134^n$ is a factor of 168&nbsp;&nbsp;$\Rightarrow$ the factor of 2 should be at least 3, for 168 to be a factor&nbsp;&nbsp;$\Rightarrow n = 3$Now,&nbsp;&nbsp;$1134^n = 1134^3 = 2^3 \times 3^{12} \times 7^3$&nbsp;&nbsp;is a factor of&nbsp;&nbsp;$168^m = \left(2^3 \times 3 \times 7\right)^m$&nbsp;&nbsp;$\Rightarrow m = 12$ as power of 3 should be at least 12.$\Rightarrow$ So, $m + n = 15$

It is given that $ a^m \cdot b^n = 144^{145} $, where $ a &gt; 1 $ and $ b &gt; 1 $.144 can be written as $ 144 = 2^4 \times 3^2 $Hence, $ a^m \cdot b^n = 144^{145} $ can be written as $ a^m \cdot b^n = \left(2^4 \times 3^2\right)^{145} = 2^{580} \times 3^{290} $We know that $ 3^{290} $ is a natural number, which implies it can be written as $ a^1 $, where $ a &gt; 1 $Hence, the least possible value of m is 1. Similarly, the largest value of n is 580.Hence, the largest value of (n-m) is (580-1) = 579The correct option is&nbsp;C

It is given that k divides m+2n and 3m+4n.Since k divides (m+2n), it implies k will also divide 3(m+2n). Therefore, k divides 3m+6n.Similarly, we know that k divides 3m+4n.We know that if two numbers a, and b both are divisible by c, then their difference (a-b) is also divisible by c.By the same logic, we can say that {(3m+6n)-(3m+4n)} is divisible by k. Hence, 2n is also divisible by k.Now, (m+2n) is divisible by k, it implies 2(m+2n) =2m+4n&nbsp;is also divisible by k.Hence, {(3m+4n)-(2m+4n)} = m is also divisible by k.Therefore, m, and 2n are also divisible by k.The correct option is B

Since there are two distinct factors other than 1, and itself, which implies the total number of factors of N is 4.It can be done in two ways.&nbsp;First case:&nbsp;N = $p^3$&nbsp;(where p is a prime number)Second case: N&nbsp;=&nbsp;$p_1$ *&nbsp;$p_2 $&nbsp;(Where&nbsp;&nbsp;$p_1$,&nbsp;$p_2 $&nbsp;are the prime numbers)From case 1, we can see that the numbers which is a cube of prime and less than 50 are 8, and 27 (2 numbers).From case 2, we will get the numbers in the form (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) {(13 numbers)}Hence, the total number of numbers having two distinct factors is (13+2) = 15.

We know that the number of factors of these two numbers is 15. We know that the factors of 15 are 1, 3, 5, and 15.The number of factors of N is&nbsp;(p+1)⋅(q+1)(Where,&nbsp;N = $a^p ⋅ b^q$,&nbsp;and a, b are prime numbers).Hence, the value of N will be least when (p+1) and (q+1) are as close as possible and a, and b are the least distinct prime numbers.Hence, p + 1 = 3 =&gt; p = 2, and q + 1 = 5 =&gt; q = 4, and the prime numbers a, and b are 2, and 3, respectively.Hence, the lowest value of N is&nbsp;N = $2^4$ ×&nbsp;$3^2$&nbsp;= $144$, and the second lowest value of N is&nbsp;N = $2^2$×&nbsp;$3^4$&nbsp;= 324Hence, the sum is (144 + 324) = 468

Given that the number of coins collected per week by two coin-collectors, A and B, are in the ratio 3: 4Let us assume A collects 3c coins per week and B collects 4c coins per week.Total number of coins collected by A in 5 weeks = 5*3c = 15c, which should be multiple of 7 =&gt; c should be multiple of 7.Total number of coins collected by B in 3 weeks = 3*4c = 12c, which should be a multiple of 24 =&gt; c should be a multiple of 2.So, the least possible value of c is lcm(2,7) = 14.Coins sold by A in a week = 3c = 3*14 = 42.

CAT Number Systems

Question 1.

Question 2.

Question 3.

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Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

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Question 16.

Free CAT Quant Questions

CAT Number Systems

Question 1.

<div><p>The sum of all real values of k for which $\left(\cfrac{1}{8}\right)^{k}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{3}}=\cfrac{1}{8}\times \left(\cfrac{1}{32768}\right)^{\cfrac{1}{k}}$, is</p></div>

Question 2.

<div><p>The sum of all four-digit numbers that can be formed with the distinct non-zero digits a, b, c, and d, with each digit appearing exactly once in every number, is 153310 + n, where n is a single digit natural number. Then, the value of (a + b + c + d + n) is</p></div>

Question 3.

<div><p>If $m$ and $n$ are natural numbers such that $n &gt; 1$, and $m^n = 2^{25} \times 3^{40}$, then $m - n$ equals</p></div>

Question 4.

<div><p>The number of all positive integers up to 500 with non-repeating digits is</p></div>

Question 5.

<div><p>Let n be the least positive integer such that 168 is a factor of $1134^n$. If m is the least positive integer such that $1134^n$ is a factor of $168^m$, then m+n equals</p></div>

Question 6.

<div><p>Let a,b,m and n be natural numbers such that a&gt;1 and b&gt;1 . If $a^m b^n=144^{145}$, then the largest possible value of n&minus;m is</p></div>

Question 7.

<div><p>For any natural numbers m,n, and k, such that k divides both m+2n and 3m+4n,k must be a common divisor of</p></div>

Question 8.

<div><p>The number of positive integers less than 50, having exactly two distinct factors other than 1 and itself, is</p></div>

Question 9.

<div><p>The sum of the first two natural numbers, each having 15 factors (including 1 and the number itself), is</p></div>

Question 10.

Question 11.

<div><p>In a 3-digit number $N$, the digits are non-zero and distinct such that none of the digits is a perfect square, and only one of the digits is a prime number. Then, the number of factors of the minimum possible value of $N$ is:</p></div>

Question 12.

<div><p>The number of divisors of $(2^{6}\times3^{5}\times5^{3}\times7^{2})$ which are of the form $(3r+1)$ where r is a non-negative integer, is</p></div>

Question 13.

<div><p>The sum of digits of the number $(625)^{65} \times (128)^{36}$, is</p></div>

Question 14.

<div><p>If $12^{12x}\times4^{24x+12}\times5^{2y}=8^{4z}\times20^{12x}\times243^{3x-6}$ where x, y and z are natural numbers, then $x+y+z$ equals</p></div>

Question 15.

<div><p>The sum of all the digits of the number $(10^{50}+10^{25}-123),$ is</p></div>

Question 16.

Free CAT Quant Questions