Question

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 $2N?$

$5$

$10$

$15$

$20$

$25$

Answer: E

Concepts:number base modular arithmetic casework

Difficulty rating: 2510

Solution:

Because $\mathrm{lcm}(25, 36, 100) = 900,$ the condition on $N$ depends only on $N \bmod 900,$ so consider $0 \le N \le 899.$ Let the last two base- $5$ digits be $a_1, a_0$ and the last two base- $6$ digits be $b_1,b_0.$ Modulo $10,$ the desired equality and $N\equiv a_0\pmod5$ force $a_0=b_0.$ The Chinese Remainder Theorem applied modulo $25$ and $36$ then gives $\begin{aligned} N&\equiv180a_1+150b_1 \\ &\quad {}+a_0\pmod{900}. \end{aligned}$ Comparing twice this residue modulo $100$ with the decimal number $10(a_1+b_1)+a_0+b_0$ reduces to $5a_1\equiv b_1\pmod{10}.$ Hence the valid pairs are exactly $(0,0),(2,0),(4,0),(1,5),(3,5).$ Each combines with $5$ choices of $a_0$ $(0 \le a_0 \le 4),$ giving $25$ values of $N.$ Thus, the correct answer is E.

2013 AMC 12B Problem 23

Answer: E

Difficulty rating: 2510

Solution:

Problem 23 in Other Years