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The sum of a two-digit number and the number obtained by reversing the order of digits is 99. If the digits differ by 3, find the number. |
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Answer» ong>Answer: The number is either 36 or 63 Step-by-step explanation: Let the digit at TENS place be x and units place be y The two digit number = 10x + y On reversing the digits, The number obtained = 10y + x Their sum = 99 10x + y + 10y + x = 99 11x + 11y = 99 11(x + y) = 99 x + y = 99/11 x + y = 9 — (1) It's also given, the digits DIFFER by 3. If digit at tens place is greater than digit at units place : x – y = 3 — (2) Add both the equations, x + y + x – y = 9 + 3 2x = 12 x = 12/2 x = 6 Tens digit = 6 Units digit = 9 – 6 = 3 Therefore, the number is 63 If digit at units place is greater than digit at tens place : y – x = 3 — (3) Add equations (1) and (3), x + y + y – x = 9 + 3 2y = 12 y = 12/2 y = 6 Units digit = 6 Tens digit = 9 – 6 = 3 Then, the number is 36 |
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