Difference between revisions of "2030 AMC 8 Problems/Problem 1"
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== Problem == | == 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? | |
| − | What is the | ||
| − | <math>\text {(A)}\ | + | <math>\text {(A)}\ 2 \qquad \text {(B)}\ 3 \qquad \text {(C)}\ 5 \qquad \text {(D)}\ 8 \qquad \text {(E)} 10</math> |
== Solution== | == Solution== | ||
Revision as of 19:22, 24 November 2024
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?
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:
![\[24x^2+25x-47=(-8x-3)(ax-2)-53\]](http://latex.artofproblemsolving.com/a/c/5/ac50cc0b6809140b1cf018b8118258ad039a453f.png)
Using the FOIL method, you should then multiply (−8𝑥−3) and (𝑎𝑥−2). After that you should have the following:
![\[24x^2+25x-47=-8ax^2-3ax+16x+6-53\]](http://latex.artofproblemsolving.com/2/4/3/243f30c8ab30b59203fbd4f570daa3870fa2fa80.png)
Then, reduce on the right side of the equation:
![\[24x^2+25x-47=-8ax^2-3ax+16x-47\]](http://latex.artofproblemsolving.com/7/2/e/72e8b252b223d6f23e50bfac60bc4b5c03a6c952.png)
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.
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See also
| 2030 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 2 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
