Difference between revisions of "2013 AMC 10B Problems/Problem 25"
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<cmath>N \equiv 5m+n \pmod{5}</cmath> | <cmath>N \equiv 5m+n \pmod{5}</cmath> | ||
<cmath>N \equiv 25m+5n \pmod{6}</cmath> | <cmath>N \equiv 25m+5n \pmod{6}</cmath> | ||
− | <cmath>2N\equiv30m + | + | <cmath>2N\equiv30m + 6n \pmod{10}</cmath> |
and this simplifies to | and this simplifies to | ||
Line 56: | Line 56: | ||
<cmath>N \equiv n \pmod{5}</cmath> | <cmath>N \equiv n \pmod{5}</cmath> | ||
<cmath>N \equiv m+5n \pmod{6}</cmath> | <cmath>N \equiv m+5n \pmod{6}</cmath> | ||
− | <cmath>2N\equiv | + | <cmath>2N\equiv 6n \pmod{10}</cmath> |
From inspection, when | From inspection, when | ||
Line 62: | Line 62: | ||
n=0, m=6 | n=0, m=6 | ||
− | n=1, m= | + | n=1, m=6 |
− | n=2, m= | + | n=2, m=2 |
− | n=3, m= | + | n=3, m=2 |
n=4, m=4 | n=4, m=4 |
Revision as of 14:37, 24 February 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
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 b=d, 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