Question

One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two digits number and add the resulting number to the original number, you get 88. What is the original number?

Answer 1

Let the tens digit be M and ones digit be N.

Therefore, Original number = 10M + N

One of the two digits of a two digit number is three times the other digit.

→ M = 3N

Now,

Original Number = 10M + N

→ 10(3N) + N

→ 31N

If you interchange the digits of this two digits number and add the resulting number to the original number, you get 88.

Interchanged number = 10N + M

Also, Interchanged number of 31N = 13N

According to question,

→ 13N + 31N = 88

→ 44N = 88

→ N = 88/44

→ N = 2

Now,

Original number = 31(2) = 62

Interchanges number = 13(2) = 26

Therefore, Original number is 62.

Answer 2

Answer:

Given:

• One of the two digits of a two digit number is three times the other digit.

• If you interchange the digits of this two digits number and add the resulting number to the original number, you get 88.

Find:

• What is the original number?

According to the question:

• Let as assume 'x' as tens digit and 'y' as ones digit.

• Let 'A' be the original number.

• Let 'B' be the incharged number.

Calculations:

⇒ A = (10x + y)

⇒ x = 3y

⇒ A = (10x + y)

⇒ 10 (3y) + y

⇒ 31y

⇒ B = (10y + x)

⇒ B = (31y = 13y)

⇒ (13y + 31y) = 88

⇒ 44y = 88

⇒ y = 88/44

⇒ y = 2

Multiplying value of ones digit with original number we get our answer.

⇒ A = (31 × 2) = 62

Now adding our results with incharged number, for verification.

⇒ B = (13 × 2) = 26

Verification:

⇒ 62 + 26 = 88

Therefore, 62 is the original number.