Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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==Solution 2 (fast)== | ==Solution 2 (fast)== | ||
Use the method from Solution <math>1</math> to get <math>63a = 27b+36c</math>. Subtract <math>27b+36c</math> from both sides to get <math>63a -27b-36c=0</math>. From here we look at possible cases. Either one (from <math>63a</math>, <math>-27b</math>, or <math>-36c</math>) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases (<math>pnn,npn,nnp,ppn,pnp,npp</math>). For all of these cases, there are 2 distinct pairs of <math>a,b</math>, and <math>c</math>, as you can simply switch the values of the 2 positives or negatives (<math>x+y=y+x</math>). This leaves us with <math>6 \cdot2 =12</math> cases. Then, we add the case where <math>a, b</math>, and <math>c</math> are all 0, giving us <math>12+1=\boxed{\textbf{(D) } 13}</math> ordered triples <math>(a,b,c).</math> | Use the method from Solution <math>1</math> to get <math>63a = 27b+36c</math>. Subtract <math>27b+36c</math> from both sides to get <math>63a -27b-36c=0</math>. From here we look at possible cases. Either one (from <math>63a</math>, <math>-27b</math>, or <math>-36c</math>) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases (<math>pnn,npn,nnp,ppn,pnp,npp</math>). For all of these cases, there are 2 distinct pairs of <math>a,b</math>, and <math>c</math>, as you can simply switch the values of the 2 positives or negatives (<math>x+y=y+x</math>). This leaves us with <math>6 \cdot2 =12</math> cases. Then, we add the case where <math>a, b</math>, and <math>c</math> are all 0, giving us <math>12+1=\boxed{\textbf{(D) } 13}</math> ordered triples <math>(a,b,c).</math> | ||
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| + | ~iluvme | ||
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== 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}} | ||
Revision as of 12:20, 14 November 2022
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 
in the infinite repeating decimal 
)
Solution 1
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
Solution 2 (fast)
Use the method from Solution 
to get 
. Subtract 
from both sides to get 
. From here we look at possible cases. Either one (from 
, 
, or 
) is positive, and the other 2 are negative, or 2 are positive, and one is negative. There are 6 of these cases (
). For all of these cases, there are 2 distinct pairs of 
, and , as you can simply switch the values of the 2 positives or negatives (
). This leaves us with 
cases. Then, we add the case where 
, and are all 0, giving us 
ordered triples
~iluvme
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. 
