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

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==Problem 24==
 
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>?
 
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>
 
<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math>
  
==Solution==
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==Solutions==
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>R=4</math> and <math>P=\boxed{\textbf{(A)}\ 1}</math>.
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===Solution 1===
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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==  
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===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>.
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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 <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>.
  
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==Video Solution==
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https://youtu.be/WJ0Hodj0h2o - Happytwin
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https://youtu.be/6xNkyDgIhEE?t=2905
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==See Also==
 
{{AMC8 box|year=2016|num-b=23|num-a=25}}
 
{{AMC8 box|year=2016|num-b=23|num-a=25}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 17:34, 21 April 2021

Problem 24

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 $P$?

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

Solutions

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

Video Solution

https://youtu.be/WJ0Hodj0h2o - Happytwin

https://youtu.be/6xNkyDgIhEE?t=2905


See Also

2016 AMC 8 (ProblemsAnswer KeyResources)
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

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