# Number System Questions for IIFT PDF

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## Number System Questions for IIFT PDF

Download important IIFT Number System Questions PDF based on previously asked questions in IIFT and other MBA exams. Practice Number System questions and answers for IIFT exam.

Practice IIFT Mock Tests

Question 1: Let N = 1421 * 1423 * 1425. What is the remainder when N is divided by 12?

a) 0

b) 9

c) 3

d) 6

Question 2: When $2^{256}$ is divided by 17, the remainder would be

a) 1

b) 16

c) 14

d) None of these

Question 3: Number S is obtained by squaring the sum of digits of a two-digit number D. If difference between S and D is 27, then the two-digit number D is

a) 24

b) 54

c) 34

d) 45

Question 4: How many 3 – digit even number can you form such that if one of the digits is 5, the following digit must be 7?

a) 5

b) 405

c) 365

d) 495

Question 5: To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: $(i_{1}, i_{2}, i_{3}$,….. are the digits of the number, starting from the most significant digit). $i_{1} i_{2} … i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + … + i_{n}.3^0$.
e.g. $259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42$
Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?

a) 2

b) 3

c) 4

d) 1

Question 6: If m and n are integers divisible by 5, which of the following is not necessarily true?

a) m – n is divisible by 5

b) m2 – n2 is divisible by 25

c) m + n is divisible by 10

d) None of these

Question 7: A certain number, when divided by 899, leaves a remainder 63. Find the remainder when the same number is divided by 29.

a) 5

b) 4

c) 1

d) Cannot be determined

Question 8: If a, b, c and d are four different positive integers selected from 1 to 25, then the highest possible value of ((a + b) + (c +d ))/((a + b) + (c – d)) would be:

a) 47

b) 49

c) 51

d) 96

e) None of the above

Question 9: Two numbers, $297_{B}$ and $792_{B}$ , belong to base B number system. If the first number is a factor of the second number then the value of B is:

a) 11

b) 12

c) 15

d) 17

e) 19

Question 10: In a Green view apartment, the houses of a row are numbered consecutively from 1 to 49. Assuming that there is a value of ‘x’ such that the sum of the numbers of the houses preceding the house numbered ‘x’ is equal to the sum of the numbers of the houses following it. Then what will be the value of ‘x’?

a) 21

b) 30

c) 35

d) 42

The numbers 1421, 1423 and 1425 when divided by 12 give remainder 5, 7 and 9 respectively.

5*7*9 mod 12 = 11 * 9 mod 12 = 99 mod 12 = 3

$2^4 = 16 = -1$ (mod $17$)
So, $2^{256} = (-1)^{64}$(mod $17$)
$= 1$ (mod $17$)
Hence, the answer is 1. Option a).

Consider the options:

24: (Square of sum of digits – the number) = 36 – 24 = 12

54: (Square of sum of digits – the number) = 81 – 54 = 27

34: (Square of sum of digits – the number) = 49 – 34 = 15

45: (Square of sum of digits – the number) = 81 – 45 = 36

So, option b) is the correct answer.

For a number to be even, its unit digit should be 0,2,4,6,8
Case 1: One of the digit is 5
Hence according to question, 5 can’t come in middle and at unit’s place, so numbers will be 570,572,574,576,578.
Case 2: No digit is 5
Hence the hundreds place can be filled in 8 ways (except 0,5) and tens place can be filled in 9 ways (except 5).
Number of ways = 8 * 9 * 5 = 360
Hence total number of ways = 360 + 5 = 365

For 203 :
first step = $2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$ = 21
second step = $2 \times 3^1 + 1 \times 3^0$ = 7
So two steps needed to reduce it to 7

Let’s say m=5k and n=5t
So m-n = 5(k-t) will be divisible by 5.
$m^2 – n^2 = 25(k^2 – t^2)$ will be divisible by 5.
$m+n = 5(k+t)$ will be divisible by 5 but not necessarily with 10.

Let’s say N is our number
N = (899K + 63) or N = ($29 \times 31$K) + 63
So when it is divided by 29, remainder will be $\frac{63}{29}$ = 5

Expression : $\frac{a + b + c + d}{a + b + c – d}$

To maximize the above expression, we have to minimize the denominator

Minimum value of the denominator = 1

So we can make $a + b + c = 26$ and $d = 25$   (as maximizing d will give denominator the least value).

So required maximum value = $\frac{a + b + c + d}{a + b + c – d}$

= $\frac{26 + 25}{26 – 25} = 51$

In Base B, $297_B = 2B^2 + 9B + 7$

and $792_B = 7B^2 + 9B + 2$

It is given that $297_{B}$ is a factor of $792_{B}$

=> $\frac{7B^2 + 9B + 2}{2B^2 + 9B + 7}$ must be an integer

=> $\frac{(2B^2 + 9B + 7) + (5B^2 – 5)}{2B^2 + 9B + 7}$

=> $\frac{5B^2 – 5}{2B^2 + 9B + 7} + 1 = k$

=> $5B^2 – 5 = (2B^2 + 9B + 7) k$      (where $k$ is factor)

Put $k = 1$

=> $5B^2 – 5 = 2B^2 + 9B + 7$

=> $B^2 – 3B – 4 = 0$

=> $(B – 4) (B + 1) = 0$

=> $B = 4 , -1$

Since, B is a base,so B must be greater than 9. Hence, it is not possible

Put $k = 2$

=> $5B^2 – 5 = 4B^2 + 18B + 14$

=> $B^2 – 18B – 19 = 0$

=> $(B – 19) (B + 1) = 0$

=> $B = 19 , -1$

$\therefore B = 19$

It is given that sum of the first $(x – 1)$ numbers is equal to sum of the numbers from $(x + 1)$ to 49
or, sum of $(x – 1)$ numbers = sum of first 49 numbers – sum of first $x$ numbers
$\dfrac{x(x – 1)}{2} = \dfrac{49 * 50}{2} – \dfrac{x(x + 1)}{2}$
On solving, we get $x$ = 35