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# The units digit of a two digit number is 3 and seven times the sum of the digits is the number itself. Find the number.

Given,

The units digit of a two digit number is 3 and seven times the sum of the digits is the number itself.

We have to find the number.

Let the number on unit place be ‘u’.

And, the number on tens place be ‘t’.

According to question,

The number on the unit place of a two-digit number is 3.

So, u = 3 --------(i)

And, it is seven times the sum of the digits of the number itself.

Required equation will be

$7(t+u) = 10t+u$ -------- (ii)

$7(t+3) = 10t+3$ {using (i) into (ii)}

$7t+21 = 10t+3$

$3t = 18$

$t = 6$.

Hence, the tens digit is 6 and the units digit is 3, so the number is 63.

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