Difference between revisions of "2016 AMC 8 Problems/Problem 24"
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<math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math> | <math>\textbf{(A) }1\qquad\textbf{(B) }2\qquad\textbf{(C) }3\qquad\textbf{(D) }4\qquad \textbf{(E) }5</math> | ||
− | ==Solution== | + | ==Video Solution== |
+ | https://youtu.be/6xNkyDgIhEE?t=2905 | ||
+ | |||
===Solution 1=== | ===Solution 1=== | ||
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>. |
Revision as of 22:40, 12 August 2020
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
?
Video Solution
https://youtu.be/6xNkyDgIhEE?t=2905
Solution 1
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
.
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, lets move on to the second piece of information that was given to us. RST is divisible by 3. So, according to the divisibility of 3 rule the sum of
has to be a multiple of 3. The only 2 big enough is 9 and 12 and since 5 is already given. The possible sums of
is 4 and 7. So, the possible values for
are 1,3,4,3 and the possible values of
is 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 know 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.
.
Video Solution
https://youtu.be/WJ0Hodj0h2o - Happytwin
2016 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 23 |
Followed by Problem 25 | |
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