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Difference between revisions of "2016 AMC 8 Problems/Problem 24"

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==Solution 2==
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We know that out of <math>PQRST</math> <math>QRS</math> is divisible by <math>5</math>. Therefore <math>S</math> is obviously 5 because <math>QRS</math> is divisible by 5. So we now have PQR5T as our number. Next, lets move on to the second piece of information that was given to us. RST is divisible by 3. So, according to the divisibility of 3 rule the sum of RST has to be a multiple of 3. The only 2 big enough is 9 and 12 and since 5 is already given. The possible sums of RT is 4 and 7. So, the possible values for R are 1,3,4,3 and the possible values of T is 3,1,3,4. So, using this we can move on to the fact that PQR is divisible by 4. So, using that we know that R has to be even so 4 is the only possible value for R. Using that we also know that 3 is the only possible value for 3. So, we know have PQRST = PQ453 so the possible values are 1 and 2 for P and Q. Using the divisibility rule of 4 we know that QR has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for P is 1.  <math>P=\boxed{\textbf{(A)}\ 1}</math>.

Revision as of 02:14, 28 October 2017

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 $P$?

$\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5$

Solution

We see that since $QRS$ is divisible by $5$, $S$ must equal either $0$ or $5$, but it cannot equal $0$, so $S=5$. We notice that since $PQR$ must be even, $R$ must be either $2$ or $4$. However, when $R=2$, we see that $T \equiv 2 \pmod{3}$, which cannot happen because $2$ and $5$ are already used up; so $R=4$. This gives $T \equiv 3 \pmod{4}$, meaning $T=3$. 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 $P=\boxed{\textbf{(A)}\ 1}$.

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png


Solution 2

We know that out of $PQRST$ $QRS$ is divisible by $5$. Therefore $S$ is obviously 5 because $QRS$ is divisible by 5. So we now have PQR5T as our number. Next, lets move on to the second piece of information that was given to us. RST is divisible by 3. So, according to the divisibility of 3 rule the sum of RST has to be a multiple of 3. The only 2 big enough is 9 and 12 and since 5 is already given. The possible sums of RT is 4 and 7. So, the possible values for R are 1,3,4,3 and the possible values of T is 3,1,3,4. So, using this we can move on to the fact that PQR is divisible by 4. So, using that we know that R has to be even so 4 is the only possible value for R. Using that we also know that 3 is the only possible value for 3. So, we know have PQRST = PQ453 so the possible values are 1 and 2 for P and Q. Using the divisibility rule of 4 we know that QR has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for P is 1. $P=\boxed{\textbf{(A)}\ 1}$.

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