# Question

A 2-digit number is equals to four times the sum of its digits. If the digits of the number are reversed, the new number formed is 27 more than the original number. By forming two simultaneous equations and solving them, find the original number.

Pls help with the above question. Thank you!

Let digit in the tens place be X and digit in the ones place be Y.

A 2-digit number is equals to four times the sum of its digits: To get the number, we multiply the first digit in the tens place by 10 and add it to Y. This will be equal to four times the sum of X and Y.

10X + Y = 4(X + Y)
10X + Y = 4X + 4Y
6X = 3Y
2X = Y
Y – 2X = 0 —– (1)

If the digits of the number are reversed, the new number formed is 27 more than the original number: To get the new number, we multiply Y by 10 since it is now in the tens place and add it to Y. This will equal the original number (10X + Y) plus an additional 27.

10Y + X = (10X + Y) + 27
9Y – 9X = 27
Y – X = 3 —– (2)

(2) – (1)

(Y – X) – (Y – 2X) = 3 – 0
X = 3

Sub x = 3 into (2):

Y – 3 = 3
Y = 6

∴ The number is 36.