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 $2N?$

$5$

$10$

$15$

$20$

$25$

Solution:

It is enough to work modulo $900=\operatorname{lcm}(25,36,100)$ , because the last two base- $5$ , base- $6$ , and decimal digits repeat with that period.

Let the last two base- $5$ digits be $a_1a_0$ and the last two base- $6$ digits be $b_1b_0$ . The units digit condition gives $a_0+b_0\equiv2N\equiv2a_0\pmod{10}$ , so $a_0=b_0$ .

Writing $N=150N_3+30a_1+a_0$ , the base- $6$ tens digit condition gives $N_3\equiv a_1+b_1\pmod6$ , so $N=900N_4+180a_1+150b_1+a_0$ .

The tens digit condition for $S$ and $2N$ reduces to $5a_1\equiv b_1\pmod{10}$ . With $0\le a_1\le4$ and $0\le b_1\le5$ , the valid pairs are $(0,0),(2,0),(4,0),(1,5),(3,5)$ .

There are $5$ choices for $a_0=b_0$ , so there are $5\cdot5=25$ choices of $N$ .

Thus, the correct answer is E .

2013 AMC 10B Problem 25

Solution:

Problem 25 in Other Years