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"

(→‎Solution)
(→‎Solution)
Line 4: Line 4:
 
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 </math>
 
<math>\textbf{(A) } 7 \qquad \textbf{(B) } 8 \qquad \textbf{(C) } 9 \qquad \textbf{(D) } 10 \qquad \textbf{(E) } 11 </math>
  
−
==Solution==
+
==Solution 1==
 
Let Joey's age be <math>j</math>, Chloe's age be <math>c</math>, and we know that Zoe's age is <math>1</math>.  
 
Let Joey's age be <math>j</math>, Chloe's age be <math>c</math>, and we know that Zoe's age is <math>1</math>.  
  
Line 12: Line 12:
  
 
-tdeng
 
-tdeng
 +
 +
==Solution 2==
 +
Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)
 +
 +
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 iff <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, <cmath>C-Z=36</cmath> <cmath>J-Z=37</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>.
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2018|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2018|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:55, 16 February 2018

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?

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

Solution 1

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 $\boxed{\text{(E) }11}$

-tdeng

Solution 2

Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)

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 iff $Z$ 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, \[C-Z=36\] \[J-Z=37\] 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 $J+n=74$. $7+4=\boxed{(\textbf{E}) 11}$.

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

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=91512"