Test: Arun Sharma Based Level 1: Remainder & Divisibility

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This mock test of Test: Arun Sharma Based Level 1: Remainder & Divisibility for CAT helps you for every CAT entrance exam. This contains 10 Multiple Choice Questions for CAT Test: Arun Sharma Based Level 1: Remainder & Divisibility (mcq) to study with solutions a complete question bank. The solved questions answers in this Test: Arun Sharma Based Level 1: Remainder & Divisibility quiz give you a good mix of easy questions and tough questions. CAT students definitely take this Test: Arun Sharma Based Level 1: Remainder & Divisibility exercise for a better result in the exam. You can find other Test: Arun Sharma Based Level 1: Remainder & Divisibility extra questions, long questions & short questions for CAT on EduRev as well by searching above.

QUESTION: 1

Let n! = 1 x 2 x 3 x……….x n for integer n ≥ 1. If p = 1! + (2 x 2!) + (3 x 3!) + ……(10 x 10!), then p + 2 when divided by 11! Leaves remainder of

A.
10
B.
0
C.
7
D.
1

Solution:

If P = 1! = 1
Then P + 2 = 3, when divided by 2! remainder will be 1.
If P = 1! + 2 × 2! = 5
Then, P + 2 = 7 when divided by 3! remainder is still 1.
Hence, P = 1! + (2 × 2!) + (3 × 3!)+ ……+ (10 × 10!)
Hence, when p + 2 is divided by 11!, the remainder is 1.
Alternative method:
P = 1 + 2 × 2! + 3 × 3! + …..10 × 10!
= (2 – 1)1! + (3 – 1)2! + (4 – 1)3! + …+ (11 – 1)10!
= 2! – 1! + 3! – 2! + …..+ 11! – 10!
= 11! – 1
Hence p + 2 = 11! + 1
Hence, when p + 2 is divided by 11!, the remainder is 1

QUESTION: 2

The remainder, when (15²³ + 23²³) is divided by 19, is:

A.
4
B.
15
C.
0
D.
18

Solution:

aⁿ+bⁿ is always divisible by a + b when n is odd.
Therefore 15²³ + 23²³ is always divisible by 15 + 23 = 38.
As 38 is a multiple of 19, 15²³ + 23²³ is divisible by 19.
Therefore,the required remainder is 0.

QUESTION: 3

What is the sum of all two-digit numbers that give a remainder of 3 when they are divided by 7?

A.
666
B.
676
C.
683
D.
777

Solution:

First of all, we have to identify such 2 digit numbers.
Obviously, they are 10, 17, 24, ….94
The required sum = 10 + 17 … 94.
Now this is an A.P. with a = 10, n = 13 and d = 7
Hence, the sum is

number system -remainder practice questions

QUESTION: 4

After the division of a number successively by 3, 4 and 7, the remainders obtained are 2,1 and 4 respectively. What will be the remainder if 84 divides the same number?

A.
80
B.
75
C.
41
D.
53

Solution:

In the successive division, the quotient of first division becomes the dividend of the second division and so on.
Let the last quotient be p, so the last dividend will be 7p + 4 which is the quotient of the second division.
So, the second dividend is (7p + 4) × 4 + 1.
Applying the same logic, the number = 3 {4(7p + 4) + 1} + 2 = 84p + 53
Hence, if the number is divided by 84, the remainder is 53.

QUESTION: 5

Let N = 55³ + 17³ – 72³. N is divisible by:

A.
both 7 and 13
B.
both 3 and 13
C.
both 17 and 7
D.
both 3 and 17

Solution:

We have N = 55³ + 17³ – 72³ = (54 + 1)³ + (18 – 1)³ – 72³
When N is divided by 3, we get remainders (1)³ + (- 1)³ – 0 = 0
Hence, the number N is divisible by 3.
Again N = (51 + 4)³ + 17³ – (68 + 4)³
When N is divided by 17, the remainder is (4)³ + 0 – (4)³ = 0
Hence, the number is divisible by 17.
Hence, the number is divisible by both 3 and 17.

QUESTION: 6

What will be the unit digit of 13⁴¹?

A.
3
B.
0
C.
1
D.
9

Solution:

As we know the last digit depends upon the unit digit of the multiplier numbers so the unit digit of 13⁴¹
is same as the last digit of 3⁴¹ and we know that the cyclicity of 3 is 4
On dividing the number 41 with 4 and we will get the remainder as 1 and the last digit will be 3¹ = 3

QUESTION: 7

What will be the unit digit when 45⁴⁵

A.
5
B.
0
C.
1
D.
9

Solution:

Here the last digit is depend upon the 5
The cyclicity of 5 is 1 so there is no need to divide the power with 1
because whatever is the power of 5 the last digit will remains 5 so the last
digit of 45⁴⁵ will be 5 .
5¹ = 5
5² = 25
5³= 125

QUESTION: 8

What will be the last digit of 3⁴x 4⁵ x 5⁶

A.
8
B.
0
C.
9
D.
3

Solution:

Last digit of 3⁴= 1
Last digit of 4⁵ = 4
Last digit of 5⁶= 5
So as we discussed earlier that the last digit depends upon the last digits
So 1 x 4 x 5 = 20 so the last digit is 0; also whenever 5 is multiplied by even
number then last digit will be 0

QUESTION: 9

Identify the last digit of (79⁴+ 87⁵)

A.
8
B.
0
C.
7
D.
4

Solution:

Last digit of 79⁴ is depends upon the the 9⁴ i.e 1
Last digit of 87⁵ is depends upon the the 7⁵i.e 7
So the last digit of the whole expression will be 1 +7 = 8
So the last digit will be 8

QUESTION: 10

What will be the last digit of 43⁵⁶ x 5⁶⁷x 45³⁴

A.
8
B.
5
C.
7
D.
4

Solution:

Last digit of 43⁵⁶= 1 because cyclicity of 3 is 4 and 56 is completely
divisible by 4 so the last term of 3⁴is 1.
Last digit of 5⁶⁷ is 5 and last digit of 45³⁴ is also 5
so that means the last digit of whole of the expression is 1 x 5 x 5 = 5 therefore the digit is 5.