During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.

2018 AMC 10B Problems/Problem 19

Revision as of 13:08, 10 August 2021 by Supercj (talk | contribs)
The following problem is from both the 2018 AMC 12B #14 and 2018 AMC 10B #19, so both problems redirect to this page.


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

Solution 2

Here's a different way of saying the above 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 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, \[C-Z=36\] \[J-Z=37\] 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 $J+n=74$. $7+4=\boxed{(\textbf{E}) 11}$

Solution 4

Denote Zoe's age with $n$, then Chloe's age is $C-1+n$ where $C$ represents Chloe's age when Zoe is one. We must have $n | C-1+n$. Obviously $n|n$, therefore, $n | C-1$ for 9 values of $n$, and therefore, $C-1$ has $9$ factors. $C-1$ either takes the form of $a^8$ (which is too large) or $a^2 b^2$. $C$ must be less than $100$ and $a$ and $b$ must be prime, therefore the only answer is $C-1 = 36$. Joey's age is $37+n$, which is divisible by $n$ when $n|37$, therefore the answer occurs when $n=37$ and Joey is $74$. $7+4=\boxed{(\textbf{E}) 11}$

Solution 5

Note that Zoe's age $n$ years from now is $n+1,$ and Chloe's age is $c+n$ for some positive integer. We want there to be $9$ nonnegative integers such that $n+1$ divides $c+n.$ In other words, \[\frac{c+n}{n+1}\] is an integer for nine values of $n.$ We can rewrite this as \[\frac{c+n}{n+1} = c - \frac{(c-1)(n)}{n+1} = c - (c-1)\frac{n}{n+1}\] We want $c-1$ to be divisible by nine values of $n-1,$ and the least possible number with nine factors is $2^2 \cdot 3^2 = 36.$ So $c-1 = 36, c = 37,$ and Joey's age is $37+1 = 38.$ We repeat the process, and we seek the second smallest $n$ for which \[\frac{37+n}{n+1} = 38 - \frac{(37)(n)}{n+1} = 38 - (37)\frac{n}{n+1}\] is an integer. This occurs for $n=36,$ and our answer is thus $38+36 = 74,$ and $7+4=\boxed{(\textbf{E}) 11}$

Video Solution



Video Solution


~ pi_is_3.14

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS