Difference between revisions of "2013 AMC 10B Problems/Problem 25"
(→Solution 3) |
(→Solution 3) |
||
Line 105: | Line 105: | ||
In addition, since <math>N</math> is a three digits integer, the only values for <math>a_1</math> are <math>a_1=0,1</math>, or otherwise <math>625a_1</math> will exceed 1000. | In addition, since <math>N</math> is a three digits integer, the only values for <math>a_1</math> are <math>a_1=0,1</math>, or otherwise <math>625a_1</math> will exceed 1000. | ||
− | Now going back to our original equation <cmath>625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4</cmath> | + | Now going back to our original equation |
+ | <cmath>625a_1+125a_2+25a_3+5a_4+a_5=216b_1+36b_2+6b_3+b_4</cmath> | ||
+ | Since <math>a_5=b_4</math>, | ||
+ | <cmath>625a_1+125a_2+25a_3+5a_4=216b_1+36b_2+6b_3</cmath> | ||
Revision as of 18:14, 7 August 2019
- 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 1
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 and equal to each other, it can be seen that , and , (otherwise and always have different parities) so , , ,
Therefore, can be written as and can be written as
Just keep in mind that can be one of five choices: or , ; 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 take 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 careful inspection, this is true when
This gives you choices for , and choices for , so the answer is
Solution 2 (Shortcut)
Notice that there are exactly possible values of . This means, in , every possible combination of digits will happen exactly once. We know that works because .
We know for sure that the units digit will add perfectly every added or subtracted, because . So we only have to care about cases of every subtracted. In each case, subtracts /adds , subtracts and adds for the 's digit.
As we can see, there are cases, including the original, that work. These are highlighted in . So, thus, there are possibilities for each case, and .
Solution 3
Notice that ranges from to digits and ranges from to digits.
Then let , denotes the digits of , , respectively such that Thus we have Now we are given Moving the to left, we have
Since , determine the unit digits of the two sides of the congruence equation, we have . Thus,
canceling out , we have Thus is a multiple of and the possibilities are .
In addition, since is a three digits integer, the only values for are , or otherwise will exceed 1000.
Now going back to our original equation Since ,
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.