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

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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 $P$ ?

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

Solution

Solution 1

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 $Q=2$ and $P=\boxed{\textbf{(A)}\ 1}$ .

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. $P=\boxed{\textbf{(A)}\ 1}$ .

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.

@@ Line 1: / Line 1: @@
+==Problem==
 The digits <math>1</math>, <math>2</math>, <math>3</math>, <math>4</math>, and <math>5</math> are each used once to write a five-digit number <math>PQRST</math>. The three-digit number <math>PQR</math> is divisible by <math>4</math>, the three-digit number <math>QRS</math> is divisible by <math>5</math>, and the three-digit number <math>RST</math> is divisible by <math>3</math>. What is <math>P</math>?
 <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
-==Solution 1==
+==Solution==
+===Solution 1===
 We see that since <math>QRS</math> is divisible by <math>5</math>, <math>S</math> must equal either <math>0</math> or <math>5</math>, but it cannot equal <math>0</math>, so <math>S=5</math>. We notice that since <math>PQR</math> must be even, <math>R</math> must be either <math>2</math> or <math>4</math>. However, when <math>R=2</math>, we see that <math>T \equiv 2 \pmod{3}</math>, which cannot happen because <math>2</math> and <math>5</math> are already used up; so <math>R=4</math>. This gives <math>T \equiv 3 \pmod{4}</math>, meaning <math>T=3</math>. Now, we see that <math>Q</math> could be either <math>1</math> or <math>2</math>, but <math>14</math> is not divisible by <math>4</math>, but <math>24</math> is. This means that <math>Q=2</math> and <math>P=\boxed{\textbf{(A)}\ 1}</math>.
-==Solution 2==
+===Solution 2===
 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 <math>PQR5T</math> 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 <math>RST</math> 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 <math>RT</math> is 4 and 7. So, the possible values for <math>R</math> are 1,3,4,3 and the possible values of <math>T</math> is 3,1,3,4. So, using this we can move on to the fact that <math>PQR</math> is divisible by 4. So, using that we know that <math>R</math> has to be even so 4 is the only possible value for <math>R</math>. Using that we also know that 3 is the only possible value for 3. So, we know have <math>PQRST</math> = <math>PQ453</math> so the possible values are 1 and 2 for <math>P</math> and <math>Q</math>. Using the divisibility rule of 4 we know that <math>QR</math> 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 <math>P</math> is 1.  <math>P=\boxed{\textbf{(A)}\ 1}</math>.
 {{AMC8 box|year=2016|num-b=23|num-a=25}}
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