Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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<math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math> | <math>\textbf{(A) } 9 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 11 \qquad \textbf{(D) } 13 \qquad \textbf{(E) } 14</math> | ||
| − | ==Solution == | + | ==Solution== |
We rewrite the given equation, then rearrange: | We rewrite the given equation, then rearrange: | ||
| Line 32: | Line 32: | ||
~MRENTHUSIASM | ~MRENTHUSIASM | ||
| − | ~Video Solution | + | ==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>. | ||
| + | |||
| + | ==Video Solution 1== | ||
| + | https://youtu.be/p6IauwE8cX8 | ||
| + | |||
| + | ==Video Solution (⚡️Lightning Fast⚡️)== | ||
| + | https://youtu.be/mgcHM0ATUks | ||
| + | |||
| + | ~Education, the Study of Everything | ||
| + | |||
| + | ==Video Solution 2== | ||
https://www.youtube.com/watch?v=YAazoVATYQA&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=4 | https://www.youtube.com/watch?v=YAazoVATYQA&list=PLmpPPbOoDfgj5BlPtEAGcB7BR_UA5FgFj&index=4 | ||
| + | |||
== See Also == | == See Also == | ||
{{AMC10 box|year=2022|ab=A|num-b=16|num-a=18}} | {{AMC10 box|year=2022|ab=A|num-b=16|num-a=18}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 20:24, 1 May 2024
Contents
Problem
How many three-digit positive integers 
are there whose nonzero digits 
and 
satisfy
![\[0.\overline{\underline{a}~\underline{b}~\underline{c}} = \frac{1}{3} (0.\overline{a} + 0.\overline{b} + 0.\overline{c})?\]](http://latex.artofproblemsolving.com/9/e/4/9e448c87f48492e9f1518a6989dee06242cab99f.png)
(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 
increases 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 1
Video Solution (⚡️Lightning Fast⚡️)
~Education, the Study of Everything
Video Solution 2
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. 
