This set of Aptitude Questions and Answers (MCQs) focuses on “Units and Tens Place Digit – Set 2”.

1. Find the units digit of the expression 25^{6785} + 36^{728} + 73^{62}.

a) 0

b) 1

c) 4

d) 6

View Answer

Explanation: The unit digit of 5

^{n}=5 and 6

^{n}=6, for all positive integers of n.

3

^{4n}=1, 3

^{4n+1}=3, 3

^{4n+2}=9, 3

^{4n+3}=7.

Hence, 3

^{62}= 3

^{4*15+2}= 9

Therefore, the units digit of the expression is 5+6+9 = 0.

2. Find the units digit of the expression 11^{1}.12^{2}.13^{3}.14^{4}.15^{5}.16^{6}.17^{7}.

a) 0

b) 2

c) 3

d) 5

View Answer

Explanation: The unit digit of 11

^{1}is 1.

The unit digit of 12

^{2}is 4.

The unit digit of 13

^{3}is 7.

The unit digit of 14

^{4}is 6.

The unit digit of 15

^{5}is 5.

The unit digit of 16

^{6}is 6.

The unit digit of 17

^{7}is 3.

The respective units place of the expression is 1*4*7*6*5*6*3 i.e., 0.

3. Find the units digit of the expression 36^{257}*33^{888}*49^{360}–64^{49}*25^{751}*31^{542}.

a) 8

b) 0

c) 6

d) 5

View Answer

Explanation: The unit digit of 36

^{257}*33

^{888}and 49

^{360}is 6, 1 and 1 respectively.

The unit digit of 36

^{257}*33

^{888}*49

^{360}is 6*1*1 i.e., 6.

The unit digit of 64

^{49}, 25

^{751}and 31

^{542}is 4, 5 and 1 respectively.

The unit digit of 64

^{49}*25

^{751}*31

^{542}is 4*5*1 i.e., 0.

The respective units place of the expression is 6 – 0 i.e., 6.

4. What is the units place digit of \(\frac{12^{75}}{3^{31}}+\frac{8^{68}}{16^{18}}\)?

a) 0

b) 1

c) 4

d) 2

View Answer

Explanation: = \(\frac{12^{75}}{3^{31}}+\frac{8^{68}}{16^{18}}\)

= \(\frac{4^{75}*3^{75}}{3^{31}} + \frac{2^{204}}{2^{72}}\) = 4

^{75}*3

^{44}+2

^{132}

Respective unit digit is 4*1 + 6 = 0.

5. Find the last two digits of 15*37*63*51*97*17*47.

a) 45

b) 35

c) 60

d) 55

View Answer

Explanation: 15*37*63*51*97*17*47 on division by 100 would give us a remainder that would be equal to its last 2 digits. First, we divide the numerator and denominator by 5 to get the expression,

= \(\frac{3*37*63*51*97*17*47}{20} \)

Using remainder theorem = \(\frac{3*17*3*11*17*17*7}{20}=\frac{51*33*289*7}{20}=\frac{11*13*9*7}{20}=\frac{143*63}{20}=\frac{3*3}{20} \)

Remainder = 9. Hence, the required remainder = 9*5 = 45, which would be the last two digits of the given number.

6. What is the unit digit of 1! + 2! + 3! + 4! + …… + 1000!?

a) 4

b) 3

c) 8

d) 6

View Answer

Explanation: In the given expression, after 5! all the values would have zero in its unit’s place. Thus, the units digit of the given expression depends on the unit’s digit of 1! + 2! + 3! + 4!

1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33

So, units digit of 1! + 2! + 3! + 4! + …… + 1000! is 3.

7. Find the unit’s digit of (45!)^{35!}?

a) 0

b) 5

c) 8

d) 2

View Answer

Explanation: 45! Is perfectly divisible by 10. Then (45!)

^{35!}should also be divisible by 10. So, unit’s place must be 0.

8. What is the unit’s digit of (7^{51!})^{^31!}?

a) 1

b) 2

c) 3

d) 4

View Answer

Explanation: 51!

^{31!}is divisible by 4 and unit’s digit of 7

^{4m}is 1. So, unit’s digit of (7

^{51!})

^{^31!}is 1.

9. Units digit of which of the following is the same as the unit digits of a^{17} + b^{17}, for any positive integer of a, b?

a) a^{2} + b^{2}

b) a^{12} + b^{12}

c) a^{13} + b^{13}

d) a^{10} + b^{10}

View Answer

Explanation: Any digit (from 0 to 9) raised to the power of the type 4k+1, where k is a natural number, will always end in the same digit. It means for all possible values of x, x

^{4k+1}has same unit digit as unit digit of x. So, a has same unit digit as unit digit of a

^{5}, a

^{9}, a

^{13}, … similarly b has same unit digit as b

^{5}, b

^{9}, b

^{13}, … So, a

^{13}+ b

^{13}has same unit digit as a

^{17}+ b

^{17}.

10. What is the digit at the hundredths place of the number (225)^{40}?

a) 6

b) 5

c) 4

d) 2

View Answer

Explanation: The last three digits of any number are the same as the remainder when we divide the number by 1000.

\(\frac{225^{40}}{1000} = \frac{5^{80}*9^{40}}{1000} = \frac{5^{77}*9^{40}}{8} \)

9

^{40}= (8+1)

^{40}gives remainder 1 when we divide it by 8.

\(\frac{5^{77}}{8} = \frac{5*(24+1)^{38}}{8}\) leaves a remainder 5.

So last three digits of 225

^{40}= 125*5 = 625.

So, the value of 100

^{th}digit is 6.

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