Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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==Remark== | ==Remark== | ||
One way to solve the Diophantine Equation, <math>7a=3b+4c</math> is by taking <math>\pmod{7}</math>, from which the equation becomes <math>0\equiv 3b-3c\pmod{7} \Longrightarrow b\equiv c\pmod{7}</math> so either <math>b=c</math> or WLOG <math>b<c, b+7=c</math>. | One way to solve the Diophantine Equation, <math>7a=3b+4c</math> is by taking <math>\pmod{7}</math>, from which the equation becomes <math>0\equiv 3b-3c\pmod{7} \Longrightarrow b\equiv c\pmod{7}</math> so either <math>b=c</math> or WLOG <math>b<c, b+7=c</math>. | ||
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+ | ==Video Solution (⚡️Lightning Fast⚡️)== | ||
+ | https://youtu.be/mgcHM0ATUks | ||
+ | |||
+ | ~Education, the Study of Everything | ||
==Video Solution 1== | ==Video Solution 1== |
Revision as of 21:50, 7 September 2023
Contents
[hide]Problem
How many three-digit positive integers are there whose nonzero digits and satisfy (The bar indicates repetition, thus is the infinite repeating decimal )
Solution
We rewrite the given equation, then rearrange: Now, this problem is equivalent to counting the ordered triples that satisfies the equation.
Clearly, the ordered triples are solutions to this equation.
The expression has the same value when:
- increases by as decreases by
- decreases by as increases by
We find more solutions from the solutions above: Note that all solutions are symmetric about
Together, we have ordered triples
~MRENTHUSIASM
Remark
One way to solve the Diophantine Equation, is by taking , from which the equation becomes so either or WLOG .
Video Solution (⚡️Lightning Fast⚡️)
~Education, the Study of Everything
Video Solution 1
https://www.youtube.com/watch?v=YAazoVATYQA&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=4
See Also
2022 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.