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"

(→‎Solution)
(→‎Solution 2)
 
(37 intermediate revisions by 19 users not shown)
Line 1: Line 1:
 +
==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==
+
==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>.
+
===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>.
 +
 
 +
~CHECKMATE2021
 +
 
 +
===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>.
 +
 
 +
~CHECKMATE2021
 +
 
 +
===Solution 3===
 +
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>.
 +
 
 +
~[[User:Bloggish|Bloggish]]
 +
 
 +
==Video Solution (CREATIVE THINKING + ANALYSIS!!!)==
 +
https://youtu.be/mDB6tbjl3fw
 +
 
 +
~Education, the Study of Everything
  
−
{{AMC8 box|year=2016|num-b=23|num-a=25}}
 
−
{{MAA Notice}}
 
  
  
 +
==Video Solution by OmegaLearn==
 +
https://youtu.be/6xNkyDgIhEE?t=2905
  
−
==Solution 2==  
+
==See Also==
−
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>.
+
{{AMC8 box|year=2016|num-b=23|num-a=25}}
 +
{{MAA Notice}}

Latest revision as of 22:18, 23 December 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

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=237699"