Difference between revisions of "2018 AMC 10B Problems/Problem 19"

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m (Solution 3)
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==Solution 2==
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==Solution 3==
 
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 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> (mathislife16).
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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==

Revision as of 00:59, 19 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

Here's a different way of saying your solution.

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. 7 + 4 = $\boxed{\text{(E) }11}$


Solution 3

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 (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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

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