2013 AMC 12B Problems/Problem 23

Problem

Bernardo chooses a three-digit positive integer $N$ 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 $S$ . For example, if $N=749$ , Bernardo writes the numbers 10,444 and 3,245, and LeRoy obtains the sum $S=13,689$ . For how many choices of $N$ are the two rightmost digits of $S$ , in order, the same as those of $2N$ ?

$\textbf{(A)}\ 5\qquad\textbf{(B)}\ 10\qquad\textbf{(C)}\ 15\qquad\textbf{(D}}\ 20\qquad\textbf{(E)}\ 25$ (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 $N \equiv a \pmod{6}$

also that $N \equiv b \pmod{5}$

After some inspection, it can be seen that b=d, and $b < 5$ , so $N \equiv a \pmod{6}$ $N \equiv a \pmod{5}$ $\implies N=a \pmod{30}$ $0 \le a \le 4$

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. $N \equiv 30x \pmod{36}$ $N \equiv 30x \equiv 5x \pmod{25}$ Both of those must add up to $2N\equiv60x \pmod{100}$

( $33 \ge x \ge 4$ )

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 :) $N \equiv 6x \equiv x \pmod{5}$ $N \equiv 5x \pmod{6}$ $2N\equiv 6x \pmod{10}$

Say that $x=5m+n$ (m is between 0-6, n is 0-4 because of constraints on x) Then

$N \equiv 5m+n \pmod{5}$ $N \equiv 25m+5n \pmod{6}$ $2N\equiv30m + 6n \pmod{10}$

and this simplifies to

$N \equiv n \pmod{5}$ $N \equiv m+5n \pmod{6}$ $2N\equiv 6n \pmod{10}$

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 $5\cdot 5 = 25 \implies \boxed{(E)}$ (If anything is incorrect please change it! Thanks!)

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

Page

Toolbox

Search

2013 AMC 12B Problems/Problem 23

Problem

Solution

See also