Question 113520
here is one more solution to this problem...


Let ten's digit be {{{x}}} and unit's digit be {{{y}}}. Then,

{{{Sum}}} of the digits = {{{x + y}}}

The number is = {{{10x + y}}}


{{{Sum}}} of the digits is {{{7}}},

{{{x + y = 7}}}…………………………………..…(1)

When the digits are reversed, the new number becomes

= {{{10y + x}}}

New number increased by 3 = 4 times the original number

 (10y + x) + 3 = 4 (10x + y)

or 10y + x + 3 = 40x + 4y

{{{10y + x - 40x - 4y = -3}}}

{{{-39x + 6y = -3}}}

{{{-13x + 2y = -1}}}…… (Dividing by 3) …(2)

{{{Multiplying}}} (1) by 2, we get,

{{{2x + 2y = 14}}}………………………………..…(3)

{{{Subtracting}}} (3) from (2),

{{{-13x + 2y = -1}}}………………..……………..(2)
{{{2x + 2y = 14}}}………………………………..…(3)

{{{-15x = - 15}}} …….=>……{{{x = 1}}}

Substituting {{{x = 1}}} in (1) we will have

{{{1 + y = 7}}} ……=>…{{{y = 6}}}

So, the number is {{{10x + y = 10 x 1 + 6 = 16}}}

	
When the digits are reversed, the new number is {{{(10y + x) = 61}}}

Check:
{{{(10y + x) + 3 = 4 (10x + y)}}}

{{{61 + 3 = 4 *16}}}

{{{64 = 64}}}