Difference between revisions of "2016 AMC 8 Problems/Problem 24"
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==Solutions== | ==Solutions== | ||
− | ===Solution 1=== | + | ===Solution 1 (Modular Arithmetic)=== |
We see that since <math>QRS</math> is divisible by <math>5</math>, <math>S</math> must equal either <math>0</math> or <math>5</math>, but it cannot equal <math>0</math>, so <math>S=5</math>. We notice that since <math>PQR</math> must be even, <math>R</math> must be either <math>2</math> or <math>4</math>. However, when <math>R=2</math>, we see that <math>T \equiv 2 \pmod{3}</math>, which cannot happen because <math>2</math> and <math>5</math> are already used up; so <math>R=4</math>. This gives <math>T \equiv 3 \pmod{4}</math>, meaning <math>T=3</math>. Now, we see that <math>Q</math> could be either <math>1</math> or <math>2</math>, but <math>14</math> is not divisible by <math>4</math>, but <math>24</math> is. This means that <math>Q=2</math> and <math>P=\boxed{\textbf{(A)}\ 1}</math>. | We see that since <math>QRS</math> is divisible by <math>5</math>, <math>S</math> must equal either <math>0</math> or <math>5</math>, but it cannot equal <math>0</math>, so <math>S=5</math>. We notice that since <math>PQR</math> must be even, <math>R</math> must be either <math>2</math> or <math>4</math>. However, when <math>R=2</math>, we see that <math>T \equiv 2 \pmod{3}</math>, which cannot happen because <math>2</math> and <math>5</math> are already used up; so <math>R=4</math>. This gives <math>T \equiv 3 \pmod{4}</math>, meaning <math>T=3</math>. Now, we see that <math>Q</math> could be either <math>1</math> or <math>2</math>, but <math>14</math> is not divisible by <math>4</math>, but <math>24</math> is. This means that <math>Q=2</math> and <math>P=\boxed{\textbf{(A)}\ 1}</math>. | ||
+ | |||
+ | ~CHECKMATE2021 | ||
===Solution 2=== | ===Solution 2=== | ||
− | We know that out of <math>PQRST,</math> <math>QRS</math> is divisible by <math>5</math>. Therefore <math>S</math> is obviously 5 because <math>QRS</math> is divisible by 5. So we now have <math>PQR5T</math> as our number. Next, | + | We know that out of <math>PQRST,</math> <math>QRS</math> is divisible by <math>5</math>. Therefore <math>S</math> is obviously 5 because <math>QRS</math> is divisible by 5. So we now have <math>PQR5T</math> as our number. Next, let's move on to the second piece of information that was given to us. <math>RST</math> is divisible by 3. So, according to the divisibility by 3 rule, the sum of <math>RST</math> has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of <math>RT</math> are 4 and 7. So, the possible values for <math>R</math> are 1,3,4,3 and the possible values of <math>T</math> are 3,1,3,4. So, using this we can move on to the fact that <math>PQR</math> is divisible by 4. So, using that we know that <math>R</math> has to be even so 4 is the only possible value for <math>R</math>. Using that we also know that 3 is the only possible value for 3. So, we have <math>PQRST</math> = <math>PQ453</math> so the possible values are 1 and 2 for <math>P</math> and <math>Q</math>. Using the divisibility rule of 4 we know that <math>QR</math> has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for <math>P</math> is 1. <math>P=\boxed{\textbf{(A)}\ 1}</math>. |
+ | |||
+ | ~CHECKMATE2021 | ||
+ | |||
+ | ==Video Solution (CREATIVE THINKING + ANALYSIS!!!)== | ||
+ | https://youtu.be/mDB6tbjl3fw | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
− | |||
− | |||
+ | ==Video Solution by OmegaLearn== | ||
https://youtu.be/6xNkyDgIhEE?t=2905 | https://youtu.be/6xNkyDgIhEE?t=2905 | ||
+ | ==See Also== | ||
{{AMC8 box|year=2016|num-b=23|num-a=25}} | {{AMC8 box|year=2016|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 10:30, 24 July 2024
Contents
Problem 24
The digits ,
,
,
, and
are each used once to write a five-digit number
. The three-digit number
is divisible by
, the three-digit number
is divisible by
, and the three-digit number
is divisible by
. What is
?
Solutions
Solution 1 (Modular Arithmetic)
We see that since is divisible by
,
must equal either
or
, but it cannot equal
, so
. We notice that since
must be even,
must be either
or
. However, when
, we see that
, which cannot happen because
and
are already used up; so
. This gives
, meaning
. Now, we see that
could be either
or
, but
is not divisible by
, but
is. This means that
and
.
~CHECKMATE2021
Solution 2
We know that out of
is divisible by
. Therefore
is obviously 5 because
is divisible by 5. So we now have
as our number. Next, let's move on to the second piece of information that was given to us.
is divisible by 3. So, according to the divisibility by 3 rule, the sum of
has to be a multiple of 3. The only 2 big enough are 9 and 12 and since 5 is already given. The possible sums of
are 4 and 7. So, the possible values for
are 1,3,4,3 and the possible values of
are 3,1,3,4. So, using this we can move on to the fact that
is divisible by 4. So, using that we know that
has to be even so 4 is the only possible value for
. Using that we also know that 3 is the only possible value for 3. So, we have
=
so the possible values are 1 and 2 for
and
. Using the divisibility rule of 4 we know that
has to be divisible by 4. So, either 14 or 24 are the possibilities, and 24 is divisible by 4. So the only value left for
is 1.
.
~CHECKMATE2021
Video Solution (CREATIVE THINKING + ANALYSIS!!!)
~Education, the Study of Everything
Video Solution by OmegaLearn
https://youtu.be/6xNkyDgIhEE?t=2905
See Also
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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 AJHSME/AMC 8 Problems and Solutions |
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