Difference between revisions of "1972 AHSME Problems/Problem 11"

Revision as of 19:43, 19 June 2019

We can rewrite the equation $47_a = 74_b$ as $4a+7 = 7b+4$ , or $7(a+b) = 11a + 3.$ Then $7(a + b) \equiv 3 \pmod{11}$ . Testing all the residues modulo 11, we find that the only solution to $7x \equiv 3 \pmod{11}$ is $x \equiv 2 \pmod{11}$ , so $a + b \equiv 2 \pmod{11}$ .

Now, since 7 is a digit in base $a$ and base $b$ , we must have $a, b \ge 8$ . We must also have $a+b \equiv 2 \pmod{11}$ , so $a+b \ge 24$ . We can have equality with $a=15, b=9$ , so the least possible value of $a+b$ is $\boxed{24}$ .

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Difference between revisions of "1972 AHSME Problems/Problem 11"

Revision as of 19:43, 19 June 2019