Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
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+ | == Solution 4 == | ||
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+ | Notice that the only values of <math>C</math> that have <math>3C = 10n+C</math> for some <math>n</math> are <math>0</math> and <math>5</math>. If <math>C=0</math>, then we have <math>AB0 \cdot 3 = 790</math>, and so <math>AB \cdot 3 = 79</math>. Notice that <math>79</math> is not divisible by <math>3</math>, so <math>C=0</math> is not a valid solution. Next, when <math>C=5</math>, we have that <math>AB5 \cdot 3 = 795</math>. Solving for <math>A</math> and <math>B</math> tells us that <math>A=2</math> and <math>B=6</math>, so the answer is <math>3 \cdot 2 + 2 \cdot 6 + 5 = 6 + 12 + 5 = \boxed{23}</math>. | ||
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+ | ~Mathdreams | ||
==See also== | ==See also== |
Revision as of 13:20, 13 July 2021
Problem
If ,
, and
each represent a single digit and they satisfy the equation
find
.
Solution
Notice that can only be
and
. However,
is not divisible by
, so
Thus,
~Bradygho
Solution 2
Clearly we see does not work, but
works with simple guess-and-check. We have
, so
and
. The answer is
~Geometry285
Solution 3
Easily, we can see that . Therefore,
We can see that
must be
or
. If
, then
This doesn't work because
isn't divisible by
. If
, then
Therefore,
. So, we have
.
- kante314 -
Solution 4
Notice that the only values of that have
for some
are
and
. If
, then we have
, and so
. Notice that
is not divisible by
, so
is not a valid solution. Next, when
, we have that
. Solving for
and
tells us that
and
, so the answer is
.
~Mathdreams
See also
The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition.