A two-digit number has digits whose product is 18. When 63 is added to the number, the digits get reversed. The number turns out to satisfy a quadratic in the tens digit. Solving by factorisation, what is the original number?

Question

Accepted Answer

29. Let tens digit a, units b, with ab = 18. Number = 10a + b; reversed = 10b + a. (10a+b)+63 = 10b+a gives b - a = 7. With ab = 18: a(a+7) = 18, a^2 + 7a - 18 = 0 = (a+9)(a-2), so a = 2, b = 9. Number = 29.

A two-digit number has digits whose product is 18. When 63 is added to the number, the digits get reversed.
The number turns out to satisfy a quadratic in the tens digit. Solving by factorisation, what is the original number?

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More Class 10 Solving by Factorisation practice

A two-digit number has digits whose product is 18. When 63 is added to the number, the digits get reversed.The number turns out to satisfy a quadratic in the tens digit. Solving by factorisation, what is the original number?

Step-by-Step Solution

Master this chapter in 30 minutes a day.

More Class 10 Solving by Factorisation practice

A two-digit number has digits whose product is 18. When 63 is added to the number, the digits get reversed.
The number turns out to satisfy a quadratic in the tens digit. Solving by factorisation, what is the original number?