Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"
(Created page with "==Problem== If <math>A</math>, <math>B</math>, and <math>C</math> each represent a single digit and they satisfy the equation <cmath>\begin{array}{cccc}& A & B & C \\ \times &...") |
(→Solution) |
||
(11 intermediate revisions by 5 users not shown) | |||
Line 3: | Line 3: | ||
==Solution== | ==Solution== | ||
+ | Notice that <math>C</math> can only be <math>0</math>, <math>1</math>, and <math>5</math>. However, <math>790</math> and <math>791</math> are not divisible by <math>3</math>, so <cmath>3 \times ABC = 795</cmath> <cmath>ABC = 265</cmath> Thus, <math>3A + 2B + C = \boxed{23}</math> | ||
+ | |||
+ | ~Bradygho | ||
+ | |||
+ | ==Solution 2== | ||
+ | Clearly we see <math>C=1</math> does not work, but <math>C=5</math> works with simple guess-and-check. We have <math>AB5=\frac{795}{3}=265</math>, so <math>A=2</math> and <math>B=6</math>. The answer is <math>3(2)+6(2)+1(5)=\boxed{23}</math> | ||
+ | |||
+ | ~Geometry285 | ||
+ | |||
+ | == Solution 3 == | ||
+ | Easily, we can see that <math>A=2</math>. Therefore,<cmath>\overline{BC} \cdot 3 = \overline{19C}.</cmath>We can see that <math>C</math> must be <math>1</math> or <math>5</math>. If <math>C=1</math>, then<cmath>\overline{B1} \cdot 3 = 191.</cmath>This doesn't work because <math>191</math> isn't divisible by <math>3</math>. If <math>C=5</math>, then<cmath>\overline{B5} \cdot 3 = 195.</cmath>Therefore, <math>B=6</math>. So, we have <math>3(2) + 2(6) + 5=6+12+5=18+5=\boxed{23}</math>. | ||
+ | |||
+ | - kante314 - | ||
+ | |||
+ | == Solution 4 == | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | ~Mathdreams | ||
+ | |||
+ | ==See also== | ||
+ | #[[2021 JMPSC Accuracy Problems|Other 2021 JMPSC Accuracy Problems]] | ||
+ | #[[2021 JMPSC Accuracy Answer Key|2021 JMPSC Accuracy Answer Key]] | ||
+ | #[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]] | ||
+ | {{JMPSC Notice}} |
Latest revision as of 10:42, 18 July 2021
Problem
If ,
, and
each represent a single digit and they satisfy the equation
find
.
Solution
Notice that can only be
,
, and
. However,
and
are 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.