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| | ==Solution== | | ==Solution== |
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| − | First, we can examine the units digits of the number base 5 and base 6 and eliminate some possibilities.
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| − | Say that <math>N \equiv a \pmod{6}</math>
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| − | also that <math>N \equiv b \pmod{5}</math>
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| − | After some inspection, it can be seen that b=d, and <math>b < 5</math>, so
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| − | <math>N \equiv a \pmod{6}</math>
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| − | <math>N \equiv a \pmod{5}</math>
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| − | <math>\implies N=a \pmod{30}</math>
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| − | <math>0 \le a \le 4 </math>
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| − | Therefore, N can be written as 30x+y
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| − | and 2N can be written as 60x+2y
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| − | Keep in mind that y can be 0, 1, 2, 3, 4, five choices;
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| − | Also, we have already found which digits of y will add up into the units digits of 2N.
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| − | 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)
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| − | 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.
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| − | <cmath>N \equiv 30x \pmod{36}</cmath>
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| − | <cmath>N \equiv 30x \equiv 5x \pmod{25}</cmath>
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| − | Both of those must add up to
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| − | <cmath>2N\equiv60x \pmod{100}</cmath>
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| − | (<math>33 \ge x \ge 4</math>)
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| − | 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
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| − | now the units digit :)
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| − | <cmath>N \equiv 6x \equiv x \pmod{5}</cmath>
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| − | <cmath>N \equiv 5x \pmod{6}</cmath>
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| − | <cmath>2N\equiv 6x \pmod{10}</cmath>
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| − | Say that <math>x=5m+n</math> (m is between 0-6, n is 0-4 because of constraints on x)
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| − | Then
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| − | <cmath>N \equiv 5m+n \pmod{5}</cmath>
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| − | <cmath>N \equiv 25m+5n \pmod{6}</cmath>
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| − | <cmath>2N\equiv30m + 5n \pmod{10}</cmath>
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| − | and this simplifies to
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| − | <cmath>N \equiv n \pmod{5}</cmath>
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| − | <cmath>N \equiv m+5n \pmod{6}</cmath>
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| − | <cmath>2N\equiv 5n \pmod{10}</cmath>
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| − | From inspection, when
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| − | n=0, m=6
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| − | n=1, m=5
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| − | n=2, m=4
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| − | n=3, m=5
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| − | n=4, m=4
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| − | This gives you 5 choices for x, and 5 choices for y, so the answer is
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| − | <math>5\cdot 5 = 25 \implies \boxed{(E)}</math>
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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 
?
