Competition problem — Algebra — Maths Olympiad Prep

Library / /21 of 31

Algebra Difficulty 5.6 AIME, harder Prove it Estonia

Juku thought of a 3-digit number that, when reversing the order of the digits, stays the same 3-digit number. Juku noticed that when adding 2016 to that number, the 4-digit number that arises is again the same 4-digit number when reading the digits from right to left. What number did Juku think of?

Solution

Let the number be $aba$ and let the number we get by adding $2016$ be $cddc$ . Clearly $c$ can only be $2$ or $3$ .

If $c = 2$ then by the ones digit the only possibility is $a = 6$ , and we have a carry from the ones to the tens digit. By the tens digit then $b + 1 + 1 = d$ or $b + 1 + 1 = d + 10$ . The second option is impossible, since by the hundreds digit we can only have $d = 6$ , if there is no carry from the tens to the hundreds digit, and $d = 7$ , if there is a carry from the tens to the hundreds digit. Thus $b + 2 = d$ . Then there is no carry to the hundreds digit, hence $d = 6$ and $b = 4$ .

If $c = 3$ then in adding the hundreds digits we must have a carry to the thousands digit which is possible only when $a = 9$ . But by the ones digit we should have $c = 5$ . The contradiction shows that this case is not possible.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.