model 4 finding unit place of a number Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (c)

Ten’s digit of original number = x

Unit’s digit = 2x

Number = 10x + 2x = 12x

According to the question,

3x – 2 = $1/6$ × 12x

3x – 2 = 2x

3x – 2x = 2

x = 2

Number = 12x = 12 × 2 = 24

Answer: (a)

Unit’s digit in 3 × 38 × 537 × 1256

= Unit’s digit in 3 × 8 × 7 × 6

= 4 × 2 = 8

Answer: (a)

Unit digit in the expansion of $25^6251$

= Unit digit in the expansion of $5^6251$ = 5

$36^528$≡ Unit digit in $6^528$ = 6

Now, $3^1 = 3; 3^2 = 9; 3^3 = 27; 3^4 = 81; 3^5 = 243 ;$....

$73^54 = 73^52 × 73^2$≡ $3^2$ = 9

Required digit = Unit’s digit of the sum 5 + 6 + 9 = 0

Answer: (a)

$7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401$

$3^1 = 3; 3^2 = 9; 3^3 = 27; 3^4 = 81;$

i.e. the digit at unit’s place gets repeated after power 4. Unit 6 remains same for any power.

∴ Required unit’s digit = Unit’s digit in the product of $7^3 × 6 × 3^1$ = 4

Answer: (c)

Unit’s digit in $3^4$ = 1

So, unit digit in $3^164$ = 1

Now, unit's digit in $(2153)^167$ = unit digit in $3^167$

= unit digit in $3^3$ = 7

Answer: (b)

Unit digit in $(264)^4$ i.e. 4 × 4 × 4 × 4 is 6

Unit digit in $(264)^100$ is also 6.

Now, $(264)^102 = (264)^100 × (264)^2$

= (Unit digit 6) × (Unit digit 6)=36

∴ Unit digit is 6

Similarly,

$(264)^103 + (264)^100 × (264)^3$

= (Unit digit 6) × (Unit digit 4)=24

∴ Unit digit is 4

Therefore, the unit digit in $(264)^102 + (264)^103$ is

6 + 4 = 10 i.e. 0.