# In a two-digit number, the sum of the digits is 1/7 of the number itself. The digits in the tens place is three more than the digit in the unit place. Find the number.

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Let the digit in tens place be `a` and digit in unit place be `b.` Now we have system of 2 equations with 2 unknowns.

`a+b=1/7(10a+b)`

`a=b+3`

Second equation is also substitution for `a` so we simply plug it into first equation.

`b+3+b=1/7(10(b+3)+b)`

`2b+3=1/7(11b+30)`

`14b+21=11b+30`

`14b-11b=30-21`

`3b=9`

`b=3`

`a=b+3`

`a=6`

**Therefore, our number is 63.**

The answer is 63.

Imagine the digits in units place is y and in ten's place is x. Then as per the question:

x = y +3 and x + y = 1/7 (10x + y)

Making substitution for x as y+ 3, we get

y + 3 + y = 1/7 (10. y + 10.3 +y)= 1/7 (11 y +30)

or 7. (2y +3) = 11y + 30

or 14 y + 21 = 11y + 30

or 3 y = 9 or y =3 and thus x = y + 3 = 6

and hence the number is **63**.