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

Answer: (a)

$4^1 = 4, 4^2 = 16, 4^3 = 64, 4^4 = 256, 4^5 = 1024$

Remainder on dividing 372 by 4 = 0

Remainder on dividing 373 by 4 = 1

Required unit digit = Unit digit of the sum of 6 + 4 = 0

Answer: (a)

$7^1 = 7, 7^2 = 49, 7^3 = 343, 7^4 = 2401, 7^5$ = 16807

i.e. The unit’s digit repeats itself after power 4.

Remainder after we divide 245 by 4 = 1

Unit’s digit in the product of $(4387)^245 × (621)^72$ = Unit’s digit in the product of $(4387)^1 × (621)^72$ = 7 × 1 = 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.

Answer: (d)

$(251)^98$ = ......1

$(21)^29$ = ....1

$(106)^100$ = ......6

$(705)^35$ = ....5

$(16)^4$ = .....6

259 = ......9

∴ Required answer = 1 + 1 – 6 + 5 – 6 + 9 = 16 – 12 = 4

Answer: (b)

$2^1 = 2, 2^2 = 4, 2^3 = 8, 2^4 = 16, 2^5 = 32$

Unit digit in the product of $(122)^173$

= Unit digit in $(122)^1$ = 2

(1 = remainder when 173 is divided by 4).

Answer: (b)

$(4)^{2m}$ gives 6 at unit digit.

$(4)^{2m +1}$ gives 4 at unit digit.

$(5)^n$ gives 5.

The same is the case with 1.

Required digit = Unit’s digit in the product of 4×5 × 1 = 0