Difference between revisions of "2013 AMC 12B Problems/Problem 23"
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also that <math>N \equiv b \pmod{5}</math> | also that <math>N \equiv b \pmod{5}</math> | ||
− | After some inspection, it can be seen that b= | + | After some inspection, it can be seen that b=a, and <math>b < 5</math>, so |
<math>N \equiv a \pmod{6}</math> | <math>N \equiv a \pmod{6}</math> | ||
<math>N \equiv a \pmod{5}</math> | <math>N \equiv a \pmod{5}</math> |
Revision as of 22:25, 10 March 2013
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 10,444 and 3,245, and LeRoy obtains the sum
. For how many choices of
are the two rightmost digits of
, in order, the same as those of
?
$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D}}\ 20\qquad\textbf{(E)}\ 25$ (Error compiling LaTeX. Unknown error_msg)
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 b=a, and , so
Therefore, N can be written as 30x+y
and 2N can be written as 60x+2y
Keep in mind that y can be 0, 1, 2, 3, 4, five choices; Also, we have already found which digits of y will add up into the units digits of 2N.
Now, examine the tens digit, x by using mod 36 and 25 to find the tens digit (units digits can be disregarded because y=0,1,2,3,4 will always work)
Then we see that N=30x+y and take it mod 25 and 36 to find the last two digits in the base 5 and 6 representation.
Both of those must add up to
()
Now, since y=0,1,2,3,4 will always work if x works, then we can treat x as a units digit instead of a tens digit in the respective bases and decrease the mods so that x 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
n=0, m=6
n=1, m=6
n=2, m=2
n=3, m=2
n=4, m=4
This gives you 5 choices for x, and 5 choices for y, so the answer is
(If anything is incorrect please change it! Thanks!)
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 |