n-25r=17 I think. I did this a while ago.
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The smallest 4 digit number divisible by 50 is 1000, so to have a remainder of 17, it is 1000 + 17 = 1017.
n + 20 - 17 + 13 - n = 20 - 17 + 13 = 3 + 13 = 16
Since 17 is a factor of 119, the remiander will be the same modulo 17. Thus, the remainder, on division by 17 would be 19 except that 19 contains one more 17 - leaving a final remainder of 2.
A remainder is the fraction left over as a result of dividing an uneven number. 52 ÷ 3 = 17 with a remainder of 1/3.
If x ≡ 39 mod 357 then: x = 357k + 39 for some integer k. Now 357 = 21×17, and 39 = 2×17+5 → x = 21×17×k + 2×17 + 5 → x = 17(21k + 2) + 5 → x = 17m + 5 where m = 21k + 2 (is an integer) → x ≡ 5 mod 17 → the remainder when the number is divided by 17 is 5.