Sum

The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number?

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#### Solution

Let us take the two-digit number such that the digit in the units place is b.

The digit in the tens place differs from b by 3.

Let us take it as b + 3.

So, the two-digit number is 10(b + 3) + b = 10b + 30 + b = 11b + 30.

With interchange of digits, the resulting two-digit number will be 10b + (b + 3) = 11b + 3.

If we add these two two-digit numbers, their sum is (11b + 30) + (11b + 3) = 11b + 11b + 30 + 3 = 22b + 33

It is given that the sum is 143.

Therefore, 22b + 33 = 143.

or 22b = 143 - 33

or 22b = 110

or b = `110/22`

or b = 5

The units digit is 5 and therefore the tens digit is 5 + 3 which is 8. The number is 85.

Is there an error in this question or solution?

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