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2021 AMC 10B Problem 13

Problem 13 of 25IntermediateAlgebraArithmetic

Let nn be a positive integer and dd be a digit such that the value of the numeral 32d‾\underline{32d} in base nn equals 263,263, and the value of the numeral 324‾\underline{324} in base nn equals the value of the numeral 11d1‾\underline{11d1} in base six. What is n+d?n + d ?

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Solution

The first statement means 3n2+2n+d=263.3n^2 + 2n+d = 263. Similarly, the second statement means 3n2+2n+43n^2 +2n +4 =63+62+6d+1= 6^3 +6^2 +6d+1 =253+6d.= 253 + 6d. Subtracting these shows us that 4−d=6d−104-d = 6d-10 7d=147d = 14 d=2.d=2. Therefore, 3n2+2n+2=263,3n^2 + 2n + 2 = 263, so n(3n+2)=261.n(3n+2) = 261. This implies n=9.n=9. Therefore, n+d=11.n+d = 11. Thus, the answer is B .
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Tagged: number base · system of equations

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