2030 AMC 8 Problems/Problem 1

Problem

There is a number 539ab that is divisible by 3 but leaves an remainder of 4 when divided by 5. The number of factors of 539ab is 36. What is the sum of ab?

$\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 8 \qquad \text {(E)} 10$

Solution

The faster way is to multiply each side of the given equation by 𝑎𝑥−2 (so you can get rid of the fraction). When you multiply each side by 𝑎𝑥−2, you should have: $24x^2+25x-47=(-8x-3)(ax-2)-53$ Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following: $24x^2+25x-47=-8ax^2-3ax+16x+6-53$ Then, reduce on the right side of the equation: $24x^2+25x-47=-8ax^2-3ax+16x-47$ Since the coefficients of the x² term has to be equal on both sides of the equation, −8a=24, and that concludes us with 𝑎=−3. $\Rightarrow\boxed{\mathrm{(B)}\ -3}$ .

2030 AMC 8 (Problems • Answer Key • Resources)
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2030 AMC 8 Problems/Problem 1

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