Difference between revisions of "2022 AMC 10A Problems/Problem 17"
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~MRENTHUSIASM | ~MRENTHUSIASM | ||
+ | ==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, and c</math>, as you can simply switch the values of the 2 positives or negatives (<math>x+y=y+x). 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).$ | ||
== 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 13:17, 14 November 2022
Problem
How many three-digit positive integers are there whose nonzero digits
and
satisfy
(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
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
, as you can simply switch the values of the 2 positives or negatives (
6 \cdot2 =12
a, b
c
12+1=\boxed{\textbf{(D) } 13}
(a,b,c).$
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 |
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