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

It is given that the units place digit is 3.  

So, let tens place digit be y.

∴ Our number = (10y + 3) …(1)

Our given condition is that seven times the sum of the digits is the number itself.

∴ By given condition,

7(y + 3) = (10 y + 3)

7 y + 21 = 10 y + 3

∴ 10 y - 7y = 21 - 3

∴ 3 y = 18

∴ y = 6

Substituting the value of y in equation1,

Number = 10 × 6 + 3 = 63

Hence, required number is 63.

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Answer 2

Answer:

63

Step-by-step explanation:

Let the digit in the 10s place = x

Digit in 1s place = 3 (given)

The two-digit number is 10x + 3

Seven times the sum of digits = 7(x + 3)

ATQ,

7(x + 3) = 10x + 3

7x + 21 = 10x + 3

3x = 18

x = 6

=> The digit in 10s place is 6

We know that, the digit in 1s place is 3

=> The number is 63

Verification:

7 times sum of digits = 7(6+3) = 7*9 = 63