a number is divisible by both 2 and 7. must it also be divisible by 14? why?

Question

another question...
a number is divisible by both 6 and 9. Must it also be divisible by 54?

Answer 1

Yes.As the number is divisible by 7 so it must be a multiple of 7.Again when the number is divisible by 2,it must also be a multiple of 2.Now being divisible both by 7 and 2,makes the number a multiple of 7*2 or 14 and hence it is divisible by 14

Answer 2

Yes, it's also divisible by 14, primarily because 2 and 7 are relatively prime.

Look at it this way. If N is divisible by 2 and by 7, you can write it as follows:

N = 2 x = 7 y. Now because N = 2 x, both N and 2 x are even numbers, so 7 y must be even. But 7 isn't even, and if y were odd, you'd have the product of two odd numbers, which is always odd.

[Want a PROOF of that? : (2m + 1)(2n + 1) is the product of two arbitrary even numbers. It's equal to 4mn + 2m + 2n + 1 = 2(2mn +m+n) +1, another odd number!]

So y must be even. Call it 2z. Then N = 7 y = 7 (2 z) = 14 z, so N is divisible by 14. QED.

Live long and prosper.

P.S. Re. your additional question. Oh, come on now, surely you can yourself think of a counter example. If a number is divisible by both 6 and 9 it CERTAINLY DOESN'T have to be divisible by 54. 18 is divisible by both 6 and 9, because it's their LCM or least common multiple. (18 = 3 x 6 = 2 x 9.) This is the kind of math you learn in ELEMENTARY SCHOOL!

It's all a question of prime factors: 6 = 2 x 3, and 9 = 3 x 3. The LCM must have 2, 3, and 3 in it to accommodate both the 6 and the 9. 2 x 3 x 3 = 18, so that anything divisible by 6 and 9 must be divisible by 18.

Answer 3

yes because 2 x 7 = 14, so if it's divisible by 2 x 7, then it follows that it also is divisible by 14. don't really know how else to explain it, but hope it helped.