Difference between revisions of "2018 AMC 10B Problems/Problem 19"
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+ | {{duplicate|[[2018 AMC 12B Problems|2018 AMC 12B #14]] and [[2018 AMC 10B Problems|2018 AMC 10B #19]]}} | ||
+ | |||
==Problem== | ==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? | + | Joey and Chloe and their daughter Zoe all have the same birthday. Joey is <math>1</math> year older than Chloe, and Zoe is exactly <math>1</math> year old today. Today is the first of the <math>9</math> 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? |
− | <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 \qquad | ||
+ | </math> | ||
==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> 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 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 | ||
+ | |||
+ | ==Solution 2== | ||
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>. | ||
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. | ||
− | 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{\ | + | 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{\textbf{(E) }11}</math>. |
+ | |||
+ | == 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 <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 | + | ==Solution 4== |
+ | 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, | |
+ | <cmath>\begin{align*} | ||
+ | C-Z&=36, \\ | ||
+ | J-Z&=37. | ||
+ | \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>. The answer is <math>7+4=\boxed{\textbf{(E) }11}</math>. | ||
− | + | ==Video Solution== | |
+ | https://youtu.be/E2SbkCQ1V84 | ||
+ | ~savannahsolver | ||
− | ==Solution | + | == Video Solution == |
− | + | https://youtu.be/zfChnbMGLVQ?t=111 | |
− | + | ~ pi_is_3.14 | |
==See Also== | ==See Also== | ||
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{{MAA Notice}} | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Introductory Number Theory Problems]] |
Latest revision as of 09:53, 6 October 2021
- The following problem is from both the 2018 AMC 12B #14 and 2018 AMC 10B #19, so both problems redirect to this page.
Contents
Problem
Joey and Chloe and their daughter Zoe all have the same birthday. Joey is year older than Chloe, and Zoe is exactly year old today. Today is the first of the 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?
Solution 1
Suppose that Chloe is years old today, so Joey is years old today. After years, Chloe and Zoe will be and years old, respectively. We are given that is an integer for nonnegative integers It follows that has positive divisors. The prime factorization of is either or Since the only possibility is from which We conclude that Joey is years old today.
Suppose that Joey's age is a multiple of Zoe's age after years, in which Joey and Zoe will be and years old, respectively. We are given that is an integer for some positive integer It follows that is divisible by so the only possibility is We conclude that Joey will be years old then, from which the answer is
~Supercj ~MRENTHUSIASM ~Zeric
Solution 2
Let Joey's age be , Chloe's age be , and we know that Zoe's age is .
We know that there must be values such that where is an integer.
Therefore, and . Therefore, we know that, as there are solutions for , there must be solutions for . We know that this must be a perfect square. Testing perfect squares, we see that , so . Therefore, . Now, since , by similar logic, , so and Joey will be and the sum of the digits is .
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 factors. Therefore, the difference between Chloe and Zoe's age is , so Chloe is , and Joey is . The common factor that will divide both of their ages is , so Joey will be . The answer is .
Solution 4
Similar approach to above, just explained less concisely and more in terms of the problem (less algebraic).
Let denote Chloe's age, denote Joey's age, and denote Zoe's age, where is the number of years from now. We are told that is a multiple of exactly nine times. Because is at and will increase until greater than , it will hit every natural number less than , including every factor of . For to be an integral multiple of , the difference must also be a multiple of , which happens if is a factor of . Therefore, has nine factors. The smallest number that has nine positive factors is . (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 and . Thus, By our above logic, the next time is a multiple of will occur when is a factor of . Because is prime, the next time this happens is at , when . The answer is .
Video Solution
~savannahsolver
Video Solution
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.