2016 AMC 8 Problems/Problem 24
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
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, let's move on to the second piece of information that was given to us. RST 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.
.
Solution 3 (Divisibility Rules)
We know that is divisible by
, so
would be either
or
. However,
is not a choice, so
. Also,
is divisible by
, so this means that
is
,
,
, or
. If
, then
has to be
or
(
is divisible by
), but both are taken. So,
.
must equal
or
, but because
,
. This leaves
~MrThinker
Video Solution
https://youtu.be/WJ0Hodj0h2o - Happytwin
https://youtu.be/6xNkyDgIhEE?t=2905
~savannahsolver
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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