Find the two-digit number such that its ones digit is one bigger than its tens digit, and the product of the desired number and the sum of digits is $616$. Answer is $56$.
2026-03-31 10:04:24.1774951464
Quadratic equations with word problems
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Say the tens digit is $A$, so the ones digit is $A+1$.
Then the number is $10A+(A+1)$, and the sum of digits is $A+(A+1)$.
You want $(10A+(A+1))(A+(A+1))=616$.
That's $(11A+1)(2A+1)=616$.
Equivalently, $22A^2+13A-615=0$.
Can you solve for $A$?
(There are two solutions, but only one is relevant.)