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

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We know that there must be <math>9</math> values <math>k\in\mathbb{Z}</math> such that <math>c+k=a(1+k)</math> where <math>a</math> is an integer.
 
We know that there must be <math>9</math> values <math>k\in\mathbb{Z}</math> such that <math>c+k=a(1+k)</math> where <math>a</math> is an integer.
  
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Therefore, <math>c-1+(1+k)=a(1+k)</math> and <math>c-1=(1+k)(a-1)</math>. Therefore, we know that, as there are <math>9</math> solutions for <math>k</math>, there must be <math>9</math> solutions for <math>c-1</math>. We know that this must be a perfect square. Testing perfect squares, we see that <math>c-1=36</math>, so <math>c=37</math>. Therefore, <math>j=38</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>37+36=83</math> and the sum of the digits is <math>\boxed{\text{(E) }11}</math>
+
Therefore, <math>c-1+(1+k)=a(1+k)</math> and <math>c-1=(1+k)(a-1)</math>. Therefore, we know that, as there are <math>9</math> solutions for <math>k</math>, there must be <math>9</math> solutions for <math>c-1</math>. We know that this must be a perfect square. Testing perfect squares, we see that <math>c-1=36</math>, so <math>c=37</math>. Therefore, <math>j=38</math>. Now, since <math>j-1=37</math>, by similar logic, <math>37=(1+k)(a-1)</math>, so <math>k=36</math> and Joey will be <math>38+36=74</math> and the sum of the digits is <math>\boxed{\text{(E) }11}</math>
  
 
-tdeng
 
-tdeng

Revision as of 15:53, 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

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

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

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