## IBPS Sample Problems On Unit Digit

Dear Reader,

Below are four problems on unit digit, you have to find the unit digit of a product or sum or power of the numbers.

**Question 1**

The unit digit of the sum 1289 + 2541 + 8215 + 6137 is:

a) 1 b) 2 c) 9 d) 3

**Answer:** b)2

Solution:

The required answer is the unit digit of the sum of the unit digits of the numbers in given sum.

i.e., The unit digit of (1289 + 2541 + 8215 + 6137)

= the unit digit of (9+1+5+7)

= the unit digit of (22)

= 2

Hence the answer is 2.

**Question 2**

The unit digit of the product 1022 x 729 x 889 x 971 is:

a) 2 b) 9 c) 1 d) 8

**Answer:** a)2

Solution:

The required answer is the unit digit of the product of the unit digits of the numbers in given product.

i.e, The unit digit of (1022 x 729 x 889 x 971)

= the unit digit of (2 x 9 x 9 x 1)

= the unit digit of (162)

= 2

Hence, the answer is 2.

**Question 3**

Find the unit digit of (7493^{263})x(151^{29})

a) 3 b) 9 c) 7 d) 1

**Answer:** c)7

Solution:

The given product is (7493^{263})x(151^{29})

Required unit digit = the unit digit of(3^{263})x(1^{29}) ...(1)

In the value of 3 to the power 4, we have the unit digit as 1.

so, we can rewrite (3^{263}) = [3^{(4x65 + 3)}] = [(3^{4})^{65}] x (3^{3})

Then from eqn(1),

The unit digit of(3^{263})x(1^{29}) = The unit digit of[(3^{4})^{65}] x (3^{3}) x 1^{29}

= The unit digit of[1^{65}] x 27 x 1

= The unit digit of 1 x 7 x 1 = 7

Hence, the answer is 7.

**Question 4**

Find the unit digit of 634^{262} + 634^{263}

a) 0 b) 1 c) 4 d) 6

**Answer:** a)0

Solution:

Given that 634^{262} + 634^{263}

= 634^{262}(1 + 634)

= (634^{262}) x 635

The unit digit of (634^{262}) x 635 = the unit digit of (4^{262} x 5)

We know that, the unit digit of 4 to the power of any odd number is 4 and the unit digit of 4 to the power of any even number is 6.

Then the unit digit of (4^{262} x 5) = unit digit of(6 x 5) = 0