Question

The units digit off two -digit number is 3 and seven times the sum of the digits is the number itself. Find the number.​

Answer 1

Answer:

unit digit is 3

tenth digit is x

then the number us 10x+3

so,

10x+3=7(x+3)

=> 10x+3=7x+21

=> 10x-7x=21-3

=> 3x=18

=> x=18/3=6

so the number is 63

Answer 2

$\huge\boxed{\mathtt\red{Solution :-}}$

Given :-

• unit place of 2-digit number is 3
• 7 times the sum of digits is the same number.

To Find :-

• the required number.

Let :-

• The required number be 10a+b

Solving :-

✒ b = 3 , b is unit place.

According to the question :-

✒ 10a+b = 7(a+b)

✒ 10a+3 = 7(a+3)

✒ 10a+3 = 7a+21

✒ 10a-7a = 21-3

✒ 3a = 18

$\large\boxed{\mathtt\green{∴a = 6}}$

Answer :-

The required number was :-

✒ 10a + b

✒ (10×6)+3

✒ 60+3

✒ 63

$\large\boxed{\mathtt\green{The\: required\: number\:is\:63}}$