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@PercyButtons3@mathstodon.xyz

2026-07-17 06:00 UTC

Problem for July 17th from the 2026 AMS Daily Epsilon of Math Calendar @DailyEpsilon@hachyderm.io #math #maths #mathematics #mathstodon

Replies (1)

  • @calx@infosec.exchange 2026-07-17 08:47

    @PercyButtons3@mathstodon.xyz @DailyEpsilon@hachyderm.io the number is equivalent to 1 mod 4, this limits what the last 2 digits of the number can be, i.e. 01, 05, 09, ..., 97 why are the last 2 digits limited, and only 2? because 4 | 100, so for any number with more than 2 digits, we can subtract enough 100s to just leave the last two digits without changing the number under mod 4 a better formalization of the above paragraph might be, any number with last digits AB, can be written as 100x + AB, with x representing rest of the digits and clearly, 100x + AB ≡ AB (mod 4) tho, i am sure this can be further improved now we just count the pairs of 2 digits, such that, we have a valid third digit to make their sum 13 examples for 01, there can be no such digit for 09, we have 4, making 409 the pairs: 05, 09, 13, 17, 25, 29, 33, 37, 41, 45, 53, 57, 61, 65, 73, 81, 93 the numbers: 805, 409, 913, 517, 625, 229, 733, 337, 841, 445, 553, 157, 661, 265, 373, 481, 193 so, count = 17

    Open ##3881001