PUMaC 2020 · 数论（A 组） · 第 3 题

PUMaC 2020 — Number Theory (Division A) — Problem 3

Find the sum (in base 10) of the three greatest numbers less than 1000 that are palindromes 10 in both base 10 and base 5.

解析

Find the sum (in base 10) of the three greatest numbers less than 1000 that are palindromes 10 in both base 10 and base 5. Proposed by: Henry Erdman Answer: 1584 4 Noting that 2 × 5 > 1000, first we consider palindromes of the form 1 XXX 1 . Such numbers 5 4 are greater than 5 = 625. Note, however, that the final digit (in base 10) must be congruent to 1 modulo 5, so the greatest palindrome in both bases is of the form 6 X 6 . Thus we have ten 10 options, and by trial and error, we find 676 = 10201 and 626 = 10001 . These are the two 10 5 10 5 largest numbers that satisfy our conditions, so we only have to find the next-largest. Note that any number greater than 4000 is also greater than 500 and thus cannot be a palindrome in 5 10 base 10 as well, since we have no number 500 < x < 625 such that the first and last digit 10 10 match and are congruent to 4 modulo 5. Similarly, for x > 3000 , we need 375 < x < 500 5 10 10 and the first and last digits of x to be congruent to 3 modulo 5. The only such palindromes are 383 and 393 , neither of which are palindromes in base 5. Moving down to the range 10 10 2000 = 250 < x < 375 , 292 = 2132 is not a palindrome in base 5, but 282 = 2112 5 10 10 10 5 10 5 is, thus we have found our third number. Summing in base 10, 676 + 626 + 282 = 1584.