Difference between revisions of "2021 JMPSC Accuracy Problems/Problem 7"

Latest revision as of 09:42, 18 July 2021

Problem

If $A$ , $B$ , and $C$ each represent a single digit and they satisfy the equation $\begin{array}{cccc}& A & B & C \\ \times & & &3 \\ \hline & 7 & 9 & C\end{array},$ find $3A+2B+C$ .

Solution

Notice that $C$ can only be $0$ , $1$ , and $5$ . However, $790$ and $791$ are not divisible by $3$ , so $3 \times ABC = 795$ $ABC = 265$ Thus, $3A + 2B + C = \boxed{23}$

~Bradygho

Solution 2

Clearly we see $C=1$ does not work, but $C=5$ works with simple guess-and-check. We have $AB5=\frac{795}{3}=265$ , so $A=2$ and $B=6$ . The answer is $3(2)+6(2)+1(5)=\boxed{23}$

~Geometry285

Solution 3

Easily, we can see that $A=2$ . Therefore, $\overline{BC} \cdot 3 = \overline{19C}.$ We can see that $C$ must be $1$ or $5$ . If $C=1$ , then $\overline{B1} \cdot 3 = 191.$ This doesn't work because $191$ isn't divisible by $3$ . If $C=5$ , then $\overline{B5} \cdot 3 = 195.$ Therefore, $B=6$ . So, we have $3(2) + 2(6) + 5=6+12+5=18+5=\boxed{23}$ .

- kante314 -

Solution 4

Notice that the only values of $C$ that have $3C = 10n+C$ for some $n$ are $0$ and $5$ . If $C=0$ , then we have $AB0 \cdot 3 = 790$ , and so $AB \cdot 3 = 79$ . Notice that $79$ is not divisible by $3$ , so $C=0$ is not a valid solution. Next, when $C=5$ , we have that $AB5 \cdot 3 = 795$ . Solving for $A$ and $B$ tells us that $A=2$ and $B=6$ , so the answer is $3 \cdot 2 + 2 \cdot 6 + 5 = 6 + 12 + 5 = \boxed{23}$ .

~Mathdreams

@@ Line 3: / Line 3: @@
 ==Solution==
-Notice that <math>C</math> can only be <math>0</math> and <math>5</math>. However, <math>790</math> is not divisible by <math>3</math>, so the number <math>ABC = 265</math>. Thus, <math>3A + 2B + C = \boxed{23}</math>
+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}}

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