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

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: (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

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: (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

Answer: (d)

Last digit of $(1001)^2008$ + 1002 = 1 + 2 = 3

Answer: (b)

Let the two–digit number be 10x + y where x < y.

Number obtained on reversing the digits =10y + x

According to the question,

10y + x = 4 (10x + y) – 24

40x + 4y – 10y – x = 24

39x – 6y = 24

13x – 2y = 8 ....(i)

Again, y – x = 7

y = x + 7 ....(ii)

13x – 2 (x + 7) = 8

13x – 2x – 14 = 8

11 x = 14 + 8 = 22

x = $22/11$ = 2

From equation (ii),

y – 2 = 7 ⇒ y = 2 + 7 = 9

Number = 10x + y =10×2+9= 29