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 "2018 AMC 10B Problems/Problem 19"

m (→‎Solution 1)
(Deleted flawed solution. It really should be at age of 74, 7+4=11. NOT at age of 38, 3+8=11.)
 
(8 intermediate revisions by 4 users not shown)
Line 13: Line 13:
  
 
==Solution 1==
 
==Solution 1==
−
Suppose that Chloe is <math>c</math> years old today, so Joey is <math>c+1</math> years old today. After <math>n</math> years, Chloe and Zoe will be <math>n+c</math> and <math>n+1</math> years old, respectively. We are given that <cmath>\frac{n+c}{n+1}=1+\frac{c-1}{n+1}</cmath> is an integer for <math>9</math> nonnegative integers <math>n.</math> It follows that <math>c-1</math> has <math>9</math> positive divisors. The prime factorization of <math>c-1</math> is either <math>p^8</math> or <math>p^2q^2.</math> Since <math>c-1<100,</math> it is clear that <math>c-1=2^2\cdot3^2=36,</math> from which <math>c=37.</math> We conclude that Joey is <math>c+1=38</math> years old today.
+
Suppose that Chloe is <math>c</math> years old today, so Joey is <math>c+1</math> years old today. After <math>n</math> years, Chloe and Zoe will be <math>n+c</math> and <math>n+1</math> years old, respectively. We are given that <cmath>\frac{n+c}{n+1}=1+\frac{c-1}{n+1}</cmath> is an integer for <math>9</math> nonnegative integers <math>n.</math> It follows that <math>c-1</math> has <math>9</math> positive divisors. The prime factorization of <math>c-1</math> is either <math>p^8</math> or <math>p^2q^2.</math> Since <math>c-1<100,</math> the only possibility is <math>c-1=2^2\cdot3^2=36,</math> from which <math>c=37.</math> We conclude that Joey is <math>c+1=38</math> years old today.
  
−
Suppose that Joey's age is a multiple of Zoe's age after <math>k</math> years, in which Joey and Zoe will be <math>k+38</math> and <math>k+1</math> years old, respectively. We are given that <cmath>\frac{k+38}{k+1}=1+\frac{37}{k+1}</cmath> is an integer for some positive integer <math>k.</math> It follows that <math>37</math> is divisible by <math>k+1,</math> so the smallest value for <math>k</math> is <math>36.</math> We conclude that Joey will be <math>k+38=74</math> years old then, from which the answer is <math>7+4=\boxed{\textbf{(E) }11}.</math>
+
Suppose that Joey's age is a multiple of Zoe's age after <math>k</math> years, in which Joey and Zoe will be <math>k+38</math> and <math>k+1</math> years old, respectively. We are given that <cmath>\frac{k+38}{k+1}=1+\frac{37}{k+1}</cmath> is an integer for some positive integer <math>k.</math> It follows that <math>37</math> is divisible by <math>k+1,</math> so the only possibility is <math>k=36.</math> We conclude that Joey will be <math>k+38=74</math> years old then, from which the answer is <math>7+4=\boxed{\textbf{(E) }11}.</math>
  
 
~Supercj ~MRENTHUSIASM ~Zeric
 
~Supercj ~MRENTHUSIASM ~Zeric
Line 30: Line 30:
 
Here's a different way of stating Solution 2:
 
Here's a different way of stating Solution 2:
  
−
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has <math>9</math> factors. Therefore, the difference between Chloe and Zoe's age is <math>36</math>, so Chloe is <math>37</math>, and Joey is <math>38</math>. The common factor that will divide both of their ages is <math>37</math>, so Joey will be <math>74</math>. <math>7 + 4 =  \boxed{\textbf{(E) }11}</math>.
+
If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has <math>9</math> factors. Therefore, the difference between Chloe and Zoe's age is <math>36</math>, so Chloe is <math>37</math>, and Joey is <math>38</math>. The common factor that will divide both of their ages is <math>37</math>, so Joey will be <math>74</math>. The answer is <math>7 + 4 =  \boxed{\textbf{(E) }11}</math>.
  
 
==Solution 4==
 
==Solution 4==
−
Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)
+
Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
  
 
Let <math>C+n</math> denote Chloe's age, <math>J+n</math> denote Joey's age, and <math>Z+n</math> denote Zoe's age, where <math>n</math> is the number of years from now. We are told that <math>C+n</math> is a multiple of <math>Z+n</math> exactly nine times. Because <math>Z+n</math> is <math>1</math> at <math>n=0</math> and will increase until greater than <math>C-Z</math>, it will hit every natural number less than <math>C-Z</math>, including every factor of <math>C-Z</math>. For <math>C+n</math> to be an integral multiple of <math>Z+n</math>, the difference <math>C-Z</math> must also be a multiple of <math>Z</math>, which happens if <math>Z</math> is a factor of <math>C-Z</math>. Therefore, <math>C-Z</math> has nine factors. The smallest number that has nine positive factors is <math>2^23^2=36</math> . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know <math>Z=1</math> and <math>J=C+1</math>. Thus,
 
Let <math>C+n</math> denote Chloe's age, <math>J+n</math> denote Joey's age, and <math>Z+n</math> denote Zoe's age, where <math>n</math> is the number of years from now. We are told that <math>C+n</math> is a multiple of <math>Z+n</math> exactly nine times. Because <math>Z+n</math> is <math>1</math> at <math>n=0</math> and will increase until greater than <math>C-Z</math>, it will hit every natural number less than <math>C-Z</math>, including every factor of <math>C-Z</math>. For <math>C+n</math> to be an integral multiple of <math>Z+n</math>, the difference <math>C-Z</math> must also be a multiple of <math>Z</math>, which happens if <math>Z</math> is a factor of <math>C-Z</math>. Therefore, <math>C-Z</math> has nine factors. The smallest number that has nine positive factors is <math>2^23^2=36</math> . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know <math>Z=1</math> and <math>J=C+1</math>. Thus,
Line 40: Line 40:
 
J-Z&=37.
 
J-Z&=37.
 
\end{align*}</cmath>
 
\end{align*}</cmath>
−
By our above logic, the next time <math>J-Z</math> is a multiple of <math>Z+n</math> will occur when <math>Z+n</math> is a factor of <math>J-Z</math>. Because <math>37</math> is prime, the next time this happens is at <math>Z+n=37</math>, when <math>J+n=74</math>. <math>7+4=\boxed{\textbf{(E) }11}</math>.
+
By our above logic, the next time <math>J-Z</math> is a multiple of <math>Z+n</math> will occur when <math>Z+n</math> is a factor of <math>J-Z</math>. Because <math>37</math> is prime, the next time this happens is at <math>Z+n=37</math>, when <math>J+n=74</math>. The answer is <math>7+4=\boxed{\textbf{(E) }11}</math>.
  
 
==Video Solution==
 
==Video Solution==
Line 47: Line 47:
 
~savannahsolver
 
~savannahsolver
  
−
== Video Solution ==
+
== Video Solution by OmegaLearn==
 
https://youtu.be/zfChnbMGLVQ?t=111
 
https://youtu.be/zfChnbMGLVQ?t=111
  

Latest revision as of 04:20, 12 April 2024

The following problem is from both the 2018 AMC 12B #14 and 2018 AMC 10B #19, so both problems redirect to this page.

Problem

Joey and Chloe and their daughter Zoe all have the same birthday. Joey is $1$ year older than Chloe, and Zoe is exactly $1$ year old today. Today is the first of the $9$ birthdays on which Chloe's age will be an integral multiple of Zoe's age. What will be the sum of the two digits of Joey's age the next time his age is a multiple of Zoe's age?

$\textbf{(A) }7 \qquad \textbf{(B) }8 \qquad \textbf{(C) }9 \qquad \textbf{(D) }10 \qquad \textbf{(E) }11 \qquad$

Solution 1

Suppose that Chloe is $c$ years old today, so Joey is $c+1$ years old today. After $n$ years, Chloe and Zoe will be $n+c$ and $n+1$ years old, respectively. We are given that \[\frac{n+c}{n+1}=1+\frac{c-1}{n+1}\] is an integer for $9$ nonnegative integers $n.$ It follows that $c-1$ has $9$ positive divisors. The prime factorization of $c-1$ is either $p^8$ or $p^2q^2.$ Since $c-1<100,$ the only possibility is $c-1=2^2\cdot3^2=36,$ from which $c=37.$ We conclude that Joey is $c+1=38$ years old today.

Suppose that Joey's age is a multiple of Zoe's age after $k$ years, in which Joey and Zoe will be $k+38$ and $k+1$ years old, respectively. We are given that \[\frac{k+38}{k+1}=1+\frac{37}{k+1}\] is an integer for some positive integer $k.$ It follows that $37$ is divisible by $k+1,$ so the only possibility is $k=36.$ We conclude that Joey will be $k+38=74$ years old then, from which the answer is $7+4=\boxed{\textbf{(E) }11}.$

~Supercj ~MRENTHUSIASM ~Zeric

Solution 2

Let Joey's age be $j$, Chloe's age be $c$, and we know that Zoe's age is $1$.

We know that there must be $9$ values $k\in\mathbb{Z}$ such that $c+k=a(1+k)$ where $a$ is an integer.

Therefore, $c-1+(1+k)=a(1+k)$ and $c-1=(1+k)(a-1)$. Therefore, we know that, as there are $9$ solutions for $k$, there must be $9$ solutions for $c-1$. We know that this must be a perfect square. Testing perfect squares, we see that $c-1=36$, so $c=37$. Therefore, $j=38$. Now, since $j-1=37$, by similar logic, $37=(1+k)(a-1)$, so $k=36$ and Joey will be $38+36=74$ and the sum of the digits is $\boxed{\textbf{(E) }11}$.

Solution 3

Here's a different way of stating Solution 2:

If a number is a multiple of both Chloe's age and Zoe's age, then it is a multiple of their difference. Since the difference between their ages does not change, then that means the difference between their ages has $9$ factors. Therefore, the difference between Chloe and Zoe's age is $36$, so Chloe is $37$, and Joey is $38$. The common factor that will divide both of their ages is $37$, so Joey will be $74$. The answer is $7 + 4 = \boxed{\textbf{(E) }11}$.

Solution 4

Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).

Let $C+n$ denote Chloe's age, $J+n$ denote Joey's age, and $Z+n$ denote Zoe's age, where $n$ is the number of years from now. We are told that $C+n$ is a multiple of $Z+n$ exactly nine times. Because $Z+n$ is $1$ at $n=0$ and will increase until greater than $C-Z$, it will hit every natural number less than $C-Z$, including every factor of $C-Z$. For $C+n$ to be an integral multiple of $Z+n$, the difference $C-Z$ must also be a multiple of $Z$, which happens if $Z$ is a factor of $C-Z$. Therefore, $C-Z$ has nine factors. The smallest number that has nine positive factors is $2^23^2=36$ . (We want it to be small so that Joey will not have reached three digits of age before his age is a multiple of Zoe's.) We also know $Z=1$ and $J=C+1$. Thus, \begin{align*} C-Z&=36, \\ J-Z&=37. \end{align*} By our above logic, the next time $J-Z$ is a multiple of $Z+n$ will occur when $Z+n$ is a factor of $J-Z$. Because $37$ is prime, the next time this happens is at $Z+n=37$, when $J+n=74$. The answer is $7+4=\boxed{\textbf{(E) }11}$.

Video Solution

https://youtu.be/E2SbkCQ1V84

~savannahsolver

Video Solution by OmegaLearn

https://youtu.be/zfChnbMGLVQ?t=111

~ pi_is_3.14

See Also

2018 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 AMC 10 Problems and Solutions
2018 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 AMC 12 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=2018_AMC_10B_Problems/Problem_19&oldid=218112"