39 x 5 = 195
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39!/34! x 5! = (39 x 38 x 37 x 36 x 35)/(5 x 4 x 3 x 2) = 575,757 of them
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.
These: 1, 3, 5, 13, 15, 39, 65, 195 (1 x 195, 3 x 65, 5 x 39, 13 x 15).
39 = 5x + 9 5x = 39 - 9 5x = 30 x = 30 / 5 x = 6
x + 7y = 39 So x = 39 - 7y Substitute for x in the second equation: 3(39 - 7y) - 2y = 2 117 - 21y - 2y = 2 115 = 23y y = 5 Substitute this value for y back into the first equation: x + 7*5 = 39 x + 35 = 39 x = 4