Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
Line 11: Line 11:
  
 
~Geometry285
 
~Geometry285
 +
 +
==See also==
 +
#[[2021 JMPSC Sprint Problems|Other 2021 JMPSC Sprint Problems]]
 +
#[[2021 JMPSC Sprint Answer Key|2021 JMPSC Sprint Answer Key]]
 +
#[[JMPSC Problems and Solutions|All JMPSC Problems and Solutions]]
 +
{{JMPSC Notice}}

Revision as of 17:21, 11 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 $0$ and $5$. However, $790$ is 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

See also

  1. Other 2021 JMPSC Sprint Problems
  2. 2021 JMPSC Sprint Answer Key
  3. All JMPSC Problems and Solutions

The problems on this page are copyrighted by the Junior Mathematicians' Problem Solving Competition. JMPSC.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_JMPSC_Accuracy_Problems/Problem_7&oldid=158055"