2016 AMC 8 Problem 24 Number Theory

 Understand the Problem

The digits $1$, $2$, $3$, $4$, and $5$ are each used once to write a five-digit number $PQRST$. The three-digit number $PQR$ is divisible by $4$, the three-digit number $QRS$ is divisible by $5$, and the three-digit number $RST$ is divisible by $3$. What is $P$? $\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$
Source of the problem

2016 AMC 8 Problem 24

Topic

Number Theory

Difficulty Level
Easy
Suggested Book
Mathematical Circles

Start with hints

Do you really need a hint? Try it first!

We see that since QRS  is divisible by $5$$S$ must equal either $0$ or $5$, but it cannot equal $0$, so $S=5$ Similarly proceed .
 We notice that since  PQR   must be even, R must be either $2$ or $4$. However, when \( R = 2 \) we see that \( T  \equiv  2  (mod \ 3) \), which cannot happen because $2$ and $5$ are already used up; so $R=4$
This gives   $T \equiv 3 \pmod{4}$    meaning $T=3$Almost we are done . Try to think the rest .
 Now, we see that Q  could be either $1$ or $2$, but $14$ is not divisible by $4$, but $24$ is. This means that $R=4$ and  the P is clearly equal to 1.

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

Problem on Series and Sequences | SMO, 2012 | Problem 23

Try this beautiful problem from Singapore Mathematics Olympiad, 2012 based on Series and Sequences. You may use sequential hints to solve the problem.

Theory of Equations | AIME 2015 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Theory of Equations.

Trigonometry Problem | AIME 2015 | Question 13

Try this beautiful problem number 13 from the American Invitational Mathematics Examination, AIME, 2015 based on Trigonometry.

Problem from Probability | AMC 8, 2004 | Problem no. 21

Try this beautiful problem from Probability from AMC 8, 2004, Problem 21.You may use sequential hints to solve the problem.

Smallest Perimeter of Triangle | AIME 2015 | Question 11

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 based on Smallest Perimeter of Triangle.

Intersection of two Squares | AMC 8, 2004 | Problem 25

Try this beautiful problem from Geometry based on Intersection of two Squares AMC-8, 2004,Problem-25. You may use sequential hints to solve the problem.

Probability | AMC 8, 2004 | Problem no. 22

Try this beautiful problem from Probability from AMC-8, 2004 Problem 22. You may use sequential hints to solve the problem.

Area of Rectangle Problem | AMC 8, 2004 | Problem 24

Try this beautiful problem from GeometryAMC-8, 2004 ,Problem-24, based on area of Rectangle. You may use sequential hints to solve the problem.

Area of Triangle | Geometry | AIME 2015 | Question 4

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2015 from Geometry based on Area of Triangle.

Probability | AIME 2015 – Question 5

Try this beautiful problem from the American Invitational Mathematics Examination, AIME, 2010 based on Probability Biased and Unbiased.