Difference between revisions of "2018 AMC 10B Problems/Problem 19"
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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 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 | + | 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>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 16:42, 17 December 2019
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
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 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 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
.
Solution 3
Similar approach to above, just explained less concisely and more in terms of the problem (less algebra-y)
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
.
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 |
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