In this article certain questions have been randomly taken from different topics of the quantitative section of CAT. Each question has been selected keeping in mind certain challenges you need to manage as a test taker. These challenges may range from concept basics to application orientation and test taking strategies.

1. 1

2. 3

3. 5

4. 7

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When y^{E (z)} is divided by z, the remainder will always be 1, where E(z) is Euler number of z and y & z are co-prime to each other. Also, when y^{E (z)k} is divided by z, where k is an integer, remainder will always be 1 i.e if the power is any multiple of the Euler number of the divisor, even in that case the remainder will be 1.

Note : The Euler number of a number x means the number of natural numbers which are less than x and are co-prime to x. Mathematically, the Euler number of a number z denoted by the symbol E(z) is calculated as E(z)= z(1- 1/p)(1- 1/q) (1- 1/r), where p, q and r are the different prime factors of z. Euler number of a prime number is always 1 less than the number.

In this case 2 and 131 are co-prime, so we can apply Euler’s Formula.

Now, the E(131)= 130 (since 131 is a prime number).

2^{1040 }can be written as 2^{(130*8)}. So the question amounts to checking the remainder of 2^{(130*8)} on divisibility by 13. Applying Euler’s Formula, the remainder is 1.

1. 10/17

2. 2/17

3. -2/5

4.None of these

Thus, r-6= 11 or r-6 = -11

r = 17, -5

mod (2q-12) = 8

Thus, 2q-12 = 8 or 2q-12 = -8

q = 10, 2

There are 4 vales which q/r can take : 10/17, -2, 2/17 and -2/5

The least of these is -2.

1. 3

2. 0

3. 1

4. 4

9/6 leaves a remainder of 3

(9+9^{2})/6 leaves a remainder of 0

(9+9^{2}+9^{3})/6 leaves a remainder of 3

(9+9^{2}+9^{3}+9^{4})/6 leaves a remainder of 0

Thus, when the expression ends in an odd power of 9 remainder is 3, and when it ends in an even power of 9, the remainder is 0. In the given case the last term ends in an odd power of 9, hence the remainder is 3.

1. 4

2. 6

3. 3

4. 7

(11^{1}-5)/8 leaves a remainder of 6

(11^{2}-5)/8 leaves a remainder of 4

(11^{3}-5)/8 leaves a remainder of 6

(11^{4}-5)/8 leaves a remainder of 4

Thus we observe that 11 raised to an odd power minus 5 leaves a remainder of 6, and 11 raised to an even power minus 5 leaves a remainder of 4. In the given case, the power of 11 is even, hence the answer is 4

**X+Y is a common prime divisor of 56 and 112.****X and Y are prime numbers**

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Thus X+Y=2 or X+Y=7

But this statement alone is not sufficient

From B, X and Y are prime numbers; which does not give any clue on their values.

Using A and B, X=Y=2 is ruled out as two prime numbers can’t add upto 2. If X+Y=7, then the only possibility is that the numbers are 2 and 5. However, which variable takes which value is not clear from here. Hence we can’t get the answer even if we use both the statements together.