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.
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.
A remainder is the fraction left over as a result of dividing an uneven number. 52 ÷ 3 = 17 with a remainder of 1/3.
"F2013" is not exactly a number. Nor is "F100".
The greatest remainder when dividing by any number is one less than the number itself. The greatest remainder when dividing by 18 is therefore 17. EXAMPLE : 35 ÷ 18 = 1 and remainder 17.
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
Oh, dude, the greatest possible remainder when you divide by 17 is 16. Like, if you divide any number by 17, the remainder can never be more than 16. So, yeah, that's the magic number you're looking for.
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.
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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.
A remainder is the fraction left over as a result of dividing an uneven number. 52 ÷ 3 = 17 with a remainder of 1/3.
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