0% Complete
0/18 Steps

2014 AMC 8 Problem 21 Divisibility (Number Theory)

Understand the problem

The $7$-digit numbers $\underline{7} \underline{4} \underline{A} \underline{5} \underline{2} \underline{B} \underline{1}$ and $\underline{3} \underline{2} \underline{6} \underline{A} \underline{B} \underline{4} \underline{C}$ are each multiples of $3$. Which of the following could be the value of $C$? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }5\qquad \textbf{(E) }8$
Source of the problem

2014 AMC 8 Problem 21

Topic

Divisibility Number Theory

Difficulty Level
Easy
Suggested Book
Mathematical Circles

Start with hints

Do you really need a hint? Try it first!

Recall the rule for divisibility by 3 . Then proceed to apply .
https://en.wikipedia.org/wiki/Divisibility_rule#Divisibility_by_3_or_9 \( \\ \) We came to know that any number N (in base 10) is divisible by 3 iff the sum of all decimal digits of the number is divisible by 3 .Alternatively any number N (in base 10) is congruent to 0 mod 3 iff the sum of all decimal digits of the number is congruent to 0 mod 3 .
So , $$ 74A52B1 \ and \ \ 326AB4C \ \ are \ multiples \ of \ 3 \\ \Rightarrow 7+4+A+5+2+B+1 \equiv 0 (mod \ 3) \ and \ 3+2+6+A+B+4+C \equiv 0 (mod \ 3) \\ \Rightarrow A+B \equiv 2 (mod \ 3) \ and A+ B+C \equiv 0 (mod \ 3) \\ $$
$$ A+B \equiv 2 (mod \ 3) \ and \ \ A+ B+C \equiv 0 (mod 3) \\ \Rightarrow C \equiv 1 (mod \ 3) $$ So , C can be 1 , 4 or 7 . So option (A) is right .

Watch the video

Connected Program at Cheenta

Math Olympiad Program

Math Olympiad is the greatest and most challenging academic contest for school students. Brilliant school students from over 100 countries participate in it every year. Cheenta works with small groups of gifted students through an intense training program. It is a deeply personalized journey toward intellectual prowess and technical sophistication.

Similar Problems

Polynomial Functional Equation – Random Olympiad Problem

This beautiful application of Functional Equation is related to the concepts of Polynomials. Sequential hints are given to work out the problem and to revisit the concepts accordingly.

Number Theory – Croatia MO 2005 Problem 11.1

This beautiful application from Croatia MO 2005, Problem 11.1 is based on the concepts of Number Theory. Sequential hints are given to work the problem accordingly.

SMO(senior)-2014 Problem 2 Number Theory

This beautiful application from SMO(senior)-2014 is based on the concepts of Number Theory . Sequential hints are provided to understand and solve the problem .

SMO (senior) -2014/problem-4 Number Theory

This beautiful application from SMO(senior)-2014/Problem 4 is based Number Theory . Sequential hints are provided to understand and solve the problem .

The best exponent for an inequality

Understand the problemLet be positive real numbers such that .Find with proof that is the minimal value for which the following inequality holds:Albania IMO TST 2013 Inequalities Medium Inequalities by BJ Venkatachala Start with hintsDo you really need a hint?...

A functional inequation

Understand the problemFind all functions such thatholds for all . Benelux MO 2013 Functional Equations Easy Functional Equations by BJ Venkatachala Start with hintsDo you really need a hint? Try it first!Note that the RHS does not contain $latex y$. Thus it should...

Mathematical Circles Inequality Problem

A beautiful inequality problem from Mathematical Circles Russian Experience . we provide sequential hints . key idea is to use arithmetic mean , geometric mean inequality.

RMO 2019

Regional Math Olympiad (RMO) 2019 is the second level Math Olympiad Program in India involving Number Theory, Geometry, Algebra and Combinatorics.

AMC 2019 12A Problem 15 Diophantine Equation

Beautiful application of Logarithm and Diophantine Equation in American Mathematics Competition (2019) 12A

Costa Rica NMO 2010, Final Round, Problem 4 – Number Theory

The problem is a beautiful application of the techniques in Diophantine Equation from Costa Rica Math Olympiad 2010.