Number system is the most fundamental topic in mathematics, one of whose subtopics include remainder theorem and unit digit. In today’s blog we will be providing you with some practice questions on remainder theorem and unit digit for SSC. First of all, let us go through the basic concepts.

Topics Covered

**Important concepts of Remainder theorem and unit digit**

**Basic remainder formula:**

Dividend = Divisor * Quotient + Remainder

If remainder = 0, then it the number is perfectly divisible by divisor.

**Note: **If we have a negative remainder (which was assumed for the convenience during solving of the problem), the divisor is to be added to get the real remainder.

**Cyclicity:**

Cyclicity is an important concept which can be used to solve questions on remainder theorem and unit digit.

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**Euler’s Remainder theorem:**

**Euler’s Remainder theorem:** For co-prime numbers M and N, Remainder [M^{E(N) }/ N] = 1, where E(N) is Euler number of N .

If N= a^{n} * b^{s} * c^{m} * d^{u} * f ^{r} such that a, b, c, d, f are prime numbers

Euler’s number of N = N (1 – 1/a)*(1 – 1/b)*(1 – 1/c)*(1 – 1/d)* (1 – 1/f)

**Remainder theorem and unit digit****: Some Important Points to Remember**

- (a
^{n }+ b^{n}) is divisible by (a + b), when n is odd. - (a
^{n }– b^{n}) is divisible by (a + b), when n is even. - (a
^{n}– b^{n}) is always divisible by (a – b), for every n. - Remainder (a*b/c) = Remainder ((Remainder(a/c) * Remainder(b/c))/c).

**Practice questions ****on remainder theorem and unit digit**

1) What is the unit digit of 1! + 2! + 3! + ……+ 88! + 89!?

(a) 5

(b) 3

(c) 1

(d) 8

Answer key: b

Solution:

1! =1

2! = 2

3! = 6

4! = 24

5! = 120

6! = 620…..

The unit digit of factorial of any number greater than 4 is 0. Hence,

Unit digit of 1! + 2! + 3! + ……+ 88! + 89! = Unit digit of 1 + 2 + 6 + 4 + 0 + 0 + ……+ 0

= Unit digit of 13 = 3

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2) If 3 divided the integer n, the remainder is 2. Then, what will be the remainder when 7n is divided by 3

(a) 3

(b) 2

(c) 6

(d) 4

Answer key: b

Solution:

Remainder (7n/3) = Remainder ((Remainder (7/3) * Remainder(n/3))/3) = Remainder (1*2/3) = 2

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3) What is the remainder when 1294*1298 is divided by 16.

(a) 14

(b) 11

(c) 12

(d) 10

Answer key: c

Solution:

16*81 =1296

Remainder (1294/16) = -2 {1294 = 16*81 – 2}

Remainder (1298/16) = 2 {1298 = 16*81 + 2}

Remainder (1294*1298/16) = -2*2 = -4 = -4+16 = 12

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4) 3^{10 }+ 5^{10} is divisible by

(a) 34

(b) 26

(c) 8

(d) 20

Answer key: a

Solution:

3^{10 }+ 5^{10} = 9^{5} + 25^{5}

We know, (a^{n }+ b^{n}) is divisible by (a + b), when n is odd.

So,

3^{10 }+ 5^{10} is divisible by 9 + 25 = 34

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5) What is remainder obtained if 455^{18} is divided by 19

(a) 0

(b) 3

(c) 4

(d) 1

Answer key: d

Solution:

Since 19 is prime, Euler’s number of 19 = 19 (1 – 1/19) = 18

Hence, by Euler’s remainder theorem, the remainder = 1

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6) What is the remainder of 1^{5}+2^{5}+ 3^{5} + 4^{5 }+ 5^{5 }+ 6^{5}+7^{5}+…..+ 50^{5} when divided by 5

(a) 3

(b) 4

(c) 2

(d) 0

Answer key: d

Solution:

When the power ‘5’ is divided by cyclicity of the numbers 0, 1, 5 and 6, the remainder = 1

When the power ‘5’ is divided by cyclicity of the numbers 2, 3, 7 and 8, the remainder = 1

When the power ‘5’ is divided by cyclicity of the numbers 4 and 9, the remainder = 1

1^{5}+2^{5}+ 3^{5} + 4^{5 }+ 5^{5 }+ 6^{5}+7^{5}+…..+ 50^{5} =1^{1}+2^{1}+ 3^{1} + 4^{1 }+ 5^{1 }+ 6^{1}+7^{1}+…..+ 50^{1} = Sum of first 50 natural numbers = 50*51/2 = 1275, divisible by 5 and hence remainder = 0

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7) What is the unit digit of 287*586*878

(a) 6

(b) 9

(c) 2

(d) 4

Answer key: a

Solution:

Unit digit of 287*586*878 = Unit digit of 7*6*8 = Unit digit of 336 = 6

Practice more to increase your speed of solving. All the best for your examination!

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