Free Numbers Set II - 02 Practice Test - CAT 

option (c)

Numbers which cannot be expressed as difference of squares are all even non-multiples of 4. There are 250 even non-multiples of 4 from 1 to 1000 and 125 such numbers between 1 and 500 (2,6,10......). 125 cannot be expressed hence 375 can be expressed. Hence option c

option (d)

First prime number =2 .To need a zero, we need a 2 and a 5 as $2\times 5=10.$ this occurs only once,

hence there will be only one zero. Answer = option d

option(b)

take $N=3$ as $\frac{3}{15}$ gives a remainder 3.

$6N=18\frac{18}{15}$ does not give a remainder 6

$7N=\frac{21}{15}$ remainder =6

$2N=\frac{6}{15}.$ remainder =6

$5N+6=\frac{21}{15}.$ Remainder = 6

Take an example of 11111111. This can be any of the 8 digits.

It would be divisible by 13 and 7 as if we could divide it into equal even groups of 3.

For 19, we need to divide it into even groups of 8. This is not true

This number is divisible by 11, where the rule of divisibility is sum of digits at odd places- sum of digits at even places.

Hence, Answer is none of these.

Ans. (c)

${74}^{288}=(37\times 2{)}^{288}$ (refer demo tutorial for last 2 digits technique)

${37}^{288}\times 2^{288}$

$({37}^4{)}^{72}\times (2{)}^{288}$

$({37}^2\times {37}^2{)}^{72}\times (\_\_56)$

$(69\times 69{)}^{72}\times (\_\_56)$

$(\_\_61{)}^{72}\times (\_\_56)$

$\left(\_\_21\right)\times \left(\_\_56\right)=\_\_76.$ answer is 7

option d
$\frac{{10}^1}{13}{\mid }_R=10$
$\frac{{10}^2}{13}{\mid }_R=9$
$\frac{{10}^3}{13}{\mid }_R=(-1)$

Hence we get ${10}^{90}{\mid }_R=1$, we are left with ${10}^1$ which will give a remainder of 10, hence the remainder will be 10 or option d.

Answer=option d

Prime factorize $24000=2^6\times 3\times 5^3$

Take out $2^6$ and then find the number of factors. This will give all the odd factors

In this case, number of factors $=(3+1)\times (1+1)=4\times 2=8$

$p+q+r+s+t+u=2005$ of the 6 integers, the number of integers that can be odd is 5,3 or 1. In the expression $(-1{)}^p+(-1{)}^q+(-1{)}^r+(-1{)}^s+(-1{)}^t+(-1{)}^u$

1)When one integer is even and the others are odd, then only one among is +1 and others are -1 each. Hence the sum of the terms is -5 + 1 = -4.

2)When three of the integers are even, then three of the terms are +1 each and the remaining three are -1 each. Hence the sum is 0.

3)When five of the integers are even, then five terms will be +1 each and the other term is -1. Hence the sum is 4. $\therefore$ The minimum value of the sum is -4.

option (c)

Any power of 10 when divided by 3 gives a remainder of 1. Also,7 divided by 3 remainder is 1. Therefore individual remainders are 1 and 1.

Thus, 1-1=0. Answer is 0

629 divided by 21 remainder is -1. Thus the question changes to $\frac{-1^{24}}{21}$ Remainder =+1

According to fermat's theorem " If P is a prime number and N is prime to P, then $N^p-N$ is divisible by P."

Open the brackets, $p^2[p^{27}-\frac{1}{p}]=p^{29}-p$ will always be divisible by 29, as it is prime.

Find the least value and the highest value of $\frac{B}{A}$

Least $\frac{B}{A}=\frac{50}{(-1)}=-50$

Highest $\frac{B}{A}=\frac{50}{0}=\infty$

$(-50,\infty )$

OPTION (B)

Observe the pattern

112358

Every $3^{rd}$ term is even. Hence a term which is a multiple of 3 will be even. 66 is a multiple of 3. Hence, the ${66}^{th}$ term is even

option (c)

$T_{n+1}=T_{n-1}+T_n.{12}^{th}$ term$=144$

${14}^{th}$ term $={12}^{th}$ term $+{13}^{th}$ term

$144+{13}^{th}$ term$=377$

${13}^{th}=233$

${15}^{th}$ term $={13}^{th}$ term $+{14}^{th}$ term $=233+377=610$

Any power of 9 when divided by 6, gives a remainder 3. hence answer $=\frac{11\times 3}{6}$((As there are $11$ terms)= $\frac{33}{6}$ = Remainder $=3$