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

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: (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)

Unit's digit in $(1570)^2$ = 0

Unit's digit in $(1571)^2$ = 1

Unit's digit in $(1572)^2$ = 4

Unit's digit in $(1573)^2$ = 9

∴ Required unit’s digit = Unit’s digit (0 +1+4 + 9) = 4

Answer: (c)

Unit’s digit in the expansion of $(22)^23$

= Unit’s digit in the expansion of $(2)^23$

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

i.e. 2 repeats itself after the index 4.

On dividing 23 by 4, remainder = 3

∴ Unit’s digit in $(2)^23$ = Unit’s digit in $(2)^3$ = 8

Answer: (a)

Expression = $(2137)^754$

Unit’s digit in 2137 = 7

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

Clearly, after index 4, the unit’s digit follow the same order.

Dividing index 754 by 4 we get remainder = 2

∴ Unit’s digit in the expansion of $(2137)^754$ = Unit’s digit in the expansion of $(2137)^2$ = 9

Answer: (a)

Ten’s digit = x

Unit’s digit = 2x – 1

Original number = 10x + (2x – 1) = 12x – 1

New number = 10 (2x – 1) + x

= 20x – 10 + x = 21x – 10

(21x – 10) – (12x + 1) = 12x – 1 – 20

9x – 9 = 12x – 21

3x = 12 ⇒ x = 4

Original number = 12x – 1 = 12 × 4 – 1 = 47