2007-06-11 22:54:38 · 3 answers · asked by reshmi20 c 1 in Science & Mathematics ➔ Mathematics
Yes it is. Here's why:
2007C1001 = 2007!/(1001!*1006!). In 2007!, there are 286 factors divisible by 7, 40 factors divisible by 49, and 5 factors divisible by 343, for a total of 331 factors of 7. Now, in 1001!, there are 143 factors divisible by 7, 20 factors divisible by 49, and 2 factors divisible by 343, for a total of 165 factors of 7. Finally, in 1006!, there are also 143 factors divisible by 7, 20 factors divisible by 49, and 2 factors divisible by 343, for a total of 165. Therefore, in the denominator, there are a total of 165+165 = 330 factors of 7. However, there are 331 factors of 7 in the numerator, so one of the factors of 7 will not be canceled. Therefore, 2007C1001 is in fact divisible by 7.
2007-06-11 23:15:45 · answer #1 · answered by Pascal 7 · 1⤊ 0⤋
Yes, it is divisible by 7. But not exactly there will be a remainder.
2007-06-12 06:02:14 · answer #2 · answered by nkellingley@btinternet.com 5 · 0⤊ 1⤋
NO. Your Hex number is not divisible by 7.
But it does have a prime number decomposition of
(227)(2803)(13513)
..In case you wanna know.
2007-06-12 06:01:21 · answer #3 · answered by Rey Arson II 3 · 0⤊ 1⤋