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

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~CHECKMATE2021
 
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===Solution 4 (Lucky and Fast)===
 
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We can simply try each of the answer choice, and we will see which one works. Trying <math> P=\boxed{\textbf{(A) }1} </math>, if <math> PQR </math> is divisible by <math> 4 </math>, <math> QR </math> must be divisible by four. Therefore, <math> QR </math> can only be <math> 24 </math>, <math> 52 </math>, or <math> 32 </math>. However, since <math> QRS </math> is divisible by <math> 5 </math>, <math> S = 5</math>, so <math> QR </math> cannot be <math> 52 </math>. When <math> QR = 32 </math>, <math> R = 2 </math>, the last requirement cannot be satisfied because <math> R + S + T = 2 + 4 + 5 = 11 </math>, and <math> 11 </math> is not divisible by <math> 3 </math>. However, when <math> QR = 24 </math>, <math> R = 4 </math>, the last requirement can be satisfied. Hence, we can see that when <math> P=\boxed{\textbf{(A) }1} </math>, there is one way to satisfy all three requirements, leading to a conclusion that <math> P </math> is <math> \boxed{\textbf{(A) }1} </math>.
 
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~[[User:CHECKMATE2021|CHECKMATE2021]]
 
  
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==

Revision as of 09:29, 24 July 2024

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 (Modular Arithmetic)

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

~CHECKMATE2021

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, let's move on to the second piece of information that was given to us. $RST$ is divisible by 3. So, according to the divisibility by 3 rule, the sum of $RST$ has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of $RT$ are 4 and 7. So, the possible values for $R$ are 1,3,4,3 and the possible values of $T$ are 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 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. $P=\boxed{\textbf{(A)}\ 1}$.

~CHECKMATE2021

Solution 3 (Divisibility Rules)

We know that $QRS$ is divisible by $5$, so $S$ would be either $5$ or $0$. However, $0$ is not a choice, so $S=5$. Also, $PQR$ is divisible by $4$, so this means that $QR$ is $12$, $32$, $24$, or $52$. If $R=2$, then $T$ has to be $2$ or $5$ ($RST$ is divisible by $3$), but both are taken. So, $R=4 \Rightarrow QR=24$. $R+S+T$ must equal $9$ or $12$, but because $4+5=9$, $R+S+T=12 \Rightarrow T=3$. This leaves $P=\boxed{\textbf{(A) }1}$

~CHECKMATE2021

Video Solution (CREATIVE THINKING + ANALYSIS!!!)

https://youtu.be/mDB6tbjl3fw

~Education, the Study of Everything


Video Solution by OmegaLearn

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

See Also

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

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