Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

m (→‎Solution 2)
(→‎Solutions)
 
(12 intermediate revisions by 6 users not shown)
Line 7: Line 7:
 
===Solution 1 (Modular Arithmetic)===
 
===Solution 1 (Modular Arithmetic)===
 
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>.
 
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>.
 +
 +
~CHECKMATE2021
  
 
===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, let's move on to the second piece of information that was given to us. <math>RST</math> is divisible by 3. So, according to the divisibility by 3 rule, the sum of <math>RST</math> 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 <math>RT</math> are 4 and 7. So, the possible values for <math>R</math> are 1,3,4,3 and the possible values of <math>T</math> are 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 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>.
 
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, let's move on to the second piece of information that was given to us. <math>RST</math> is divisible by 3. So, according to the divisibility by 3 rule, the sum of <math>RST</math> 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 <math>RT</math> are 4 and 7. So, the possible values for <math>R</math> are 1,3,4,3 and the possible values of <math>T</math> are 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 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>.
  
−
===Solution 3 (Divisibility Rules)===
+
~CHECKMATE2021
−
We know that <math>QRS</math> is divisible by <math>5</math>, so <math>S</math> would be either <math>5</math> or <math>0</math>. However, <math>0</math> is not a choice, so <math>S=5</math>. Also, <math>PQR</math> is divisible by <math>4</math>, so this means that <math>QR</math> is <math>12</math>, <math>32</math>, <math>24</math>, or <math>52</math>. If <math>R=2</math>, then <math>T</math> has to be <math>2</math> or <math>5</math> (<math>RST</math> is divisible by <math>3</math>), but both are taken. So, <math>R=4 \Rightarrow QR=24</math>. <math>R+S+T</math> must equal <math>9</math> or <math>12</math>, but because <math>4+5=9</math>, <math>R+S+T=12 \Rightarrow T=3</math>. This leaves <math>P=\boxed{\textbf{(A) }1}</math>
 
  
−
~MrThinker
+
===Solution 3 (Lucky and Fast)===
 +
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>.
  
−
==Video Solution==
+
~[[User:Bloggish|Bloggish]]
−
https://youtu.be/WJ0Hodj0h2o - Happytwin
+
 
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/mDB6tbjl3fw
 +
 
 +
~Education, the Study of Everything
  
−
https://youtu.be/6xNkyDgIhEE?t=2905
 
  
−
https://youtu.be/4eKdjtugZUA
 
  
−
~savannahsolver
+
==Video Solution by OmegaLearn==
 +
https://youtu.be/6xNkyDgIhEE?t=2905
  
 
==See Also==
 
==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}}

Latest revision as of 06:53, 27 August 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 (Lucky and Fast)

We can simply try each of the answer choice, and we will see which one works. Trying $P=\boxed{\textbf{(A) }1}$, if $PQR$ is divisible by $4$, $QR$ must be divisible by four. Therefore, $QR$ can only be $24$, $52$, or $32$. However, since $QRS$ is divisible by $5$, $S = 5$, so $QR$ cannot be $52$. When $QR = 32$, $R = 2$, the last requirement cannot be satisfied because $R + S + T = 2 + 4 + 5 = 11$, and $11$ is not divisible by $3$. However, when $QR = 24$, $R = 4$, the last requirement can be satisfied. Hence, we can see that when $P=\boxed{\textbf{(A) }1}$, there is one way to satisfy all three requirements, leading to a conclusion that $P$ is $\boxed{\textbf{(A) }1}$.

~Bloggish

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

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2016_AMC_8_Problems/Problem_24&oldid=225851"