Question

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

Answer 1

Given:

unit digit of a two digit number = 3

seven times the sum of digit is the number itself

To Find:

the two digit number

Solution:

we know that the unit place is 3

So, consider the ten's place as "y"

$⟹(10y + 3) \: \: \: \: \: \: \: \: \: ......(1)$

From the question,

seven times the sum of the digits is the number itself,

From the above condition:

$⟹(y + 3) \: \: \: \: \: \: \: \: \: \: \: \: .....(2)$

Combining equation (1) and (2),we get:

$⟹7(y + 3) = 10(y + 3)$

$⟹7y + 21 = 10y + 3$

$⟹7y - 10y = 3 - 21$

$⟹ - 3y = - 18$

$⟹$$\tt\underline\red{y\:=\:6}$

Substituting the value of "y" in 1 equation we get:

$⟹10y + 3$

$⟹10(6) + 3 = 60 + 3 = 63$

Hence:

∴ ${\boxed{\sf{\red{the\: required\: number\:is\:63}}}}$