**Q) **A number consists of two digits. Where the number is divided by the sum of its digits, the quotient is 7. If 27 is subtracted from the number, the digits interchange their places, find the number

**Ans: **Let’s consider X and Y are the digits of the given number.

Hence the given number is 10 X + Y

∴ the sum of the digits = X + Y

By 1st given condition:

∴ = 7

∴ (10 X + Y) = 7 ( X + Y)

∴ 3 X = 6 Y

∴ X = 2 Y ……….. (i)

Next, by other given condition, when 27 is subtracted from original number, teh digits interchange.

∴ we will get new number as: 10 Y + X

∴ (10 X + Y) – 27 = (10 Y + X)

9 X = 9 Y + 27

X = Y + 3 …………… (ii)

Next, putting value of X from equation (i) into equation (ii), we get:

(2Y) = Y + 3

∴ Y = 3

from equation (i), we have X = 2Y

∴ X = 2 (3) = 6

Hence, the original number: 10 X + Y = 10 (6) + 3 = 63.

**Therefore, the given number is 63.**

**Check:**

*1) Sum of the digits is 6 + 3 = 9 and if we divide 63 by 9, we get 7*

*2) 63 – 27 = 36 i.e digits interchange the position.*

*Hence our solution is correct.*