Difference between revisions of "2013 AMC 10B Problems/Problem 25"
Mathcool2009 (talk | contribs) m (→Solution) |
m (→See also) |
||
Line 73: | Line 73: | ||
== See also == | == See also == | ||
{{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}} | ||
− | + | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 23:46, 22 November 2013
- The following problem is from both the 2013 AMC 12B #23 and 2013 AMC 10B #25, so both problems redirect to this page.
Problem
Bernardo chooses a three-digit positive integer and writes both its base-5 and base-6 representations on a blackboard. Later LeRoy sees the two numbers Bernardo has written. Treating the two numbers as base-10 integers, he adds them to obtain an integer
. For example, if
, Bernardo writes the numbers
and
, and LeRoy obtains the sum
. For how many choices of
are the two rightmost digits of
, in order, the same as those of
?
Solution
First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
Say that
also that
Substituting these equations into the question and setting the units digits of 2N and S equal to each other, it can be seen that , and
, so
,
,
,
Therefore, can be written as
and
can be written as
Keep in mind that can be
, five choices;
Also, we have already found which digits of
will add up into the units digits of
.
Now, examine the tens digit, by using
and
to find the tens digit (units digits can be disregarded because
will always work)
Then we see that
and take it
and
to find the last two digits in the base
and
representation.
Both of those must add up to
()
Now, since will always work if
works, then we can treat
as a units digit instead of a tens digit in the respective bases and decrease the mods so that
is now the units digit.
Say that (m is between 0-6, n is 0-4 because of constraints on x)
Then
and this simplifies to
From inspection, when
This gives you choices for
, and
choices for
, so the answer is
See also
2013 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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.