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Slovenia — National Math Olympiad

Number theory Difficulty 5.1 AIME, harder Prove it Slovenia

Let $a$ , $b$ and $c$ be non-zero digits. Let $p$ be a prime number which divides the three-digit numbers $abc$ and $cba$ . Show that $p$ divides at least one of the numbers $a+b+c$ , $a-b+c$ and $a-c$ .

Solution

Since the prime number $p$ divides the numbers $\overline{abc} = 100a + 10b + c$ and $\overline{cba} = 100c + 10b + a$ , it must also divide their difference
$\overline{abc} - \overline{cba} = 100(a-c) + (c-a) = 99(a-c).$
If $p$ divides $a-c$ we are done. If not, then $p$ divides $99$ , so $p=3$ or $p=11$ .

If $p=3$ then $\overline{abc}$ is divisible by $3$ . A number is divisible by $3$ if and only if the sum of its digits is divisible by $3$ , so $3$ divides $a+b+c$ . Hence, $p$ divides $a+b+c$ .

If $p=11$ then $11$ divides
$\overline{abc} = 100a + 10b + c = 99a + 11b + a - b + c = 11(9a + b) + (a - b + c),$
which implies that $a-b+c$ is divisible by $11$ . In this case $p$ divides $a-b+c$ .

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