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 256785 + 36728 + 7362.
a) 0
b) 1
c) 4
d) 6
View Answer
Explanation: The unit digit of 5n=5 and 6n=6, for all positive integers of n.
34n=1, 34n+1=3, 34n+2=9, 34n+3=7.
Hence, 362 = 34*15+2 = 9
Therefore, the units digit of the expression is 5+6+9 = 0.
2. Find the units digit of the expression 111.122.133.144.155.166.177.
a) 0
b) 2
c) 3
d) 5
View Answer
Explanation: The unit digit of 111 is 1.
The unit digit of 122 is 4.
The unit digit of 133 is 7.
The unit digit of 144 is 6.
The unit digit of 155 is 5.
The unit digit of 166 is 6.
The unit digit of 177 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 36257*33888*49360–6449*25751*31542.
a) 8
b) 0
c) 6
d) 5
View Answer
Explanation: The unit digit of 36257*33888 and 49360 is 6, 1 and 1 respectively.
The unit digit of 36257*33888*49360 is 6*1*1 i.e., 6.
The unit digit of 6449, 25751 and 31542 is 4, 5 and 1 respectively.
The unit digit of 6449*25751*31542 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: = \(\frac{12^{75}}{3^{31}}+\frac{8^{68}}{16^{18}}\)
= \(\frac{4^{75}*3^{75}}{3^{31}} + \frac{2^{204}}{2^{72}}\) = 475*344+2132
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 (751!)^31!?
a) 1
b) 2
c) 3
d) 4
View Answer
Explanation: 51!31! is divisible by 4 and unit’s digit of 74m is 1. So, unit’s digit of (751!)^31! is 1.
9. Units digit of which of the following is the same as the unit digits of a17 + b17, for any positive integer of a, b?
a) a2 + b2
b) a12 + b12
c) a13 + b13
d) a10 + b10
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, x4k+1 has same unit digit as unit digit of x. So, a has same unit digit as unit digit of a5, a9, a13, … similarly b has same unit digit as b5, b9, b13, … So, a13 + b13 has same unit digit as a17 + b17.
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.
\(\frac{225^{40}}{1000} = \frac{5^{80}*9^{40}}{1000} = \frac{5^{77}*9^{40}}{8} \)
940 = (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 22540 = 125*5 = 625.
So, the value of 100th digit is 6.
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