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# Unit Digit Questions for SSC CGL

Download SSC CGL Questions on Unit Degit PDF based on previous papers very useful for SSC CGL exams. Unit Degit Questions for SSC exams.

Question 1: The unit digit in the product $122^{173}$ is

a) 2

b) 4

c) 6

d) 8

Question 2: The unit digit in the sum of (124)372 + (124)373 is

a) 5

b) 4

c) 20

d) 0

Question 3: The digit in the unit place in the square root of 66049 is

a) 3

b) 7

c) 8

d) 2

Question 4: Find the unit digit in the product $(4387)^{245} \times (621)^{72}$.

a) 1

b) 2

c) 5

d) 7

Question 5: If in a two digit number, the digit at unit place is z and the digit at tens place is 8, then the number is

a) 80z + z

b) 80 + z

c) 8z + 8

d) 80z + 1

Question 6: Find the unit place digit in 71 x 72 x 73 x 74 x 76 x 77 x 78 x 79.

a) 2

b) 0

c) 4

d) 6

Question 7: What is the unit digit of the sum of first 111 whole numbers?

a) 4

b) 6

c) 5

d) 0

Question 8: What is the unit digit of $(217)^{413} \times (819)^{547} \times (414)^{624} \times (342)^{812}$?

a) 2

b) 4

c) 6

d) 8

Question 9: If the unit digit of $(433 \times 456 \times 43N)$ is $(N + 2)$, then what is the value of $N$?

a) 1

b) 8

c) 3

d) 6

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power
as power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2.

Both of numbers have unit digit as 4 and it has a repeating cycle of 2 with unit digits as 4 and 6
so in first number power is 372 which is exactly divisible by 2 hence unit digit of first number will be 6.
and in second number power is 373 which exceeds one with the reapeating cycle of 2 hence its unit digit will be 4.

now unit digit of the sum will be 6+4 = 10

Square root of 66049 = 257

Thus, unit’s digit = 7

we need to find unit digit of $(4387)^{245} \times (621)^{72}$

unit digit of ${4387^{245}}$ = unit digit of ${7^1}$ = 7

unit digit of ${621^{72}}$ = 1

and hence 7 x 1 = 7 is the unit digit for the given expression

Digit at unit’s place = z

Digit at ten’s place = 8

=> 2-digit number = $(10 \times 8) + (1 \times z)$

= 80 + z

=> Ans – (B)

Expression : 71 x 72 x 73 x 74 x 76 x 77 x 78 x 79

Unit place is the product of unit digits.

= $(1\times2\times3\times4)\times(6\times7)\times(8\times9)$

= $24\times42\times72$

$\equiv4\times2\times2=16$

Thus, unit digit = 6

=> Ans – (D)

Sum of first 11 whole numbers is 0+1+2…110
i.e n(n+1)/2 =110*111/2
=55*111
Therefore units digit is 5

Power series of 7 i.e units digit 7 power expansion has 7,9,3 and 1 and it is raised to power 413 i.e 413/4 remainder 1 and so last digit is 7
Power series of 9 i.e units digit 9 power expansion has 9 and 1 and it is raised to power 547 i.e 547/2 remainder 1 and so last digit is 9
Power series of 4 i.e units digit 4 power expansion has 4 and 6 and it is raised to power 624 i.e 624/2 remainder 0 and so last digit is 6
Power series of 2 i.e units digit 2 power expansion has 2,4,8 and 6 and it is raised to power 812 i.e 812/4 remainder 0 and so last digit is 6
All the last digits product=7*9*6*6
=8

So,Unit digit of  $8N=N+2$ .