Question

The sum of the digits of a 2-digit number is 7. If the digits are reversed, the new number increased by 3 equals
4 times the original number. Find the original number.​

Answer 1

Answer:

Here sum of two digits equals to 7

possible two digits are = (1,6), (2,5),(3,4)

When the digits are reversed and increased by 3

new numbers formed

61 = 61+3 = 64

52= 52+3=55

43 = 43+3= 46

Now 4 times original number

16 *4 = 64

25*4=100

43*4 = 172

Hence cleary original number is (1,6) that is 16

Answer 2

Let x be the digit at ten's place and y be the digit at unit place.

$\sf \therefore \: The \: number \: = \: 10 \: x + y$

$\sf \: Sum \: of \: its \: digits = x + y$

$\sf \: On \: reversing \: the \: digits,$

$\sf \: The \: number \: becomes \: 10 \: y + x.$

$\sf \: A/Q, \\ \sf \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x + y = 7 \\ \sf \: and, \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: 10y + x + 3 = 4(10x + y) \\ \\ \sf \red{x + y = 7 }\: \: \: \: \: \\ \sf \: 10y + x + 3 = 4(10x + y) \\ \sf \implies \: 10y + x + 3 = 40x + 4y \\ \sf \implies \: 10y - 4y + x - 40x = 3 \\ \sf \implies \red{6y - 39x = 3} \: \: \: \: \blue{} \\ \\ \\ \sf \: x + y = 7 \: \: \red{\times 6} \\ \sf \: 6x + 6y = 42 \: \: \: \: \blue{eq.(i)}\\ \\ \sf \: 6y - 39x = 3 \: \: \red{ \times 1} \\ \sf \: 6y - 39x = 3 \: \: \blue{...eq.(ii)} \\ \\ \sf \: on \: solving \: we \: get \: - \\ \sf \: x = 1 \: \: \: \: \: \: \: \: y = 6 \\ \\ \sf \therefore \: Required \: number = 10x + y \\ \sf \: = \: 10 \times 1 + 6 \\ \boxed{ \red{ \underline{ \sf \: = 16 \: \: ....(ans) }}}$