Question

The sum of the digits of a 2-digit number is 10. The number obtained by interchangeing the digits exceeds the original number by 36. Find the original number​

Answer 1

$x+y=10\\10y+x=10x+y+36\\\\x+y=10\\9x-9y=-36\\\\x+y=10\\\underline{x-y=-4}\\2x=6\\x=3\\\\3+y=10\\y=7\\\\10\cdot3+7=37$

37

Answer 2

$\huge\mathfrak{\underline{\underline{Question:}}}$

The sum of the digits of a 2-digit number is 10. The number obtained by interchangeing the digits exceeds the original number by 36. Find the original number

_____________________

$\underline{Let\: the -}$

• ten's digit be M

• one's digit be N

$\huge\mathfrak{\underline{\underline{Solution:}}}$

∴ Number = 10M + N

Sum of two digit number is 10.

$\implies$ M + N = 10

$\implies$M = 10 - N ...(1)

(The number obtained by interchanging the digits exceeds the original number by 36.)

∴ Interchanged number = 10N + M

$\implies$10N + M = 10M + N +36

$\implies$10N - N + M - 10M = 36

$\implies$9N - 9M = 36

$\implies$N - M = 4

$\implies$N - (10 - N) = 4 [From (1)]

$\implies$N - 10 + N = 4

$\implies$2N = 14

$\implies$N = 7

Substitute value of N in (1)

$\implies$M = 10 -7

$\implies$ M = 3

•°• Original number = 10M + N

$\implies$10(3) + 7

$\implies$30 + 7

$\huge\bold{=37}$