Difference between revisions of "2013 AMC 10B Problems/Problem 25"
(→See also) |
(→See also) |
||
Line 74: | Line 74: | ||
{{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}} | {{AMC12 box|year=2013|ab=B|num-b=22|num-a=24}} | ||
{{AMC10 box|year=2013|ab=B|num-b=24|after=Last question}} | {{AMC10 box|year=2013|ab=B|num-b=24|after=Last question}} | ||
+ | {{MAA Notice}} |
Revision as of 14:04, 3 July 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
After some inspection, 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 |
2013 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last question | |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.