Question

Prove: 3/ a(2a^2 + 7)

Answer 1

Huh?

Answer 2

Yes, you are to prove that 3 divides [(a)(2a^2+7)] probably by using the greatest common divisor.

Answer 3

Sorry, can not help

Answer 4

There must be more. For instance, is a is 1, 3 is not a divisor, if a=0, 3 is not a divisor.

Answer 5

the exact problem states:
for an arbitrary integer a, verify that 3/a(2a^2+7). This is a problem out of a number theory book included in the section involving gcd.

Answer 6

actually a = 1 does work
a(2*1+7) = 9

a = 2
2(2*4+7) = 30

a = 3
3(2*9+7) obviously works but 75

a=4
4(2*16+7) = 4*39 etc
however I do not know how to do the proof

Answer 7

try recurssion ?
(a+1)(2(a+1)^2+7)
(a+1)(2 a^2 + 4 a + 9)
= 2 a^3 + 6 a^2 + 13 a + 9
subtract original 2a^3 +7a
and get
6 a^2 + 6 a + 9
so
the difference between each successive value of a is divisible by 3

Answer 8

Is that form of proof allowed?

Answer 9

You can also work Mod 3:

a(2 a^2 + 7) = a(-a^2 + 1)

Then, if a = 0, the expression is zero. Else, we have that by Fermat's little theorem that a^2 = 1. So, working Mod 3, the expression is always equal to zero, which means that the original expression (not reduced Mod 3) is always divisible by 3.

Answer 10

thanks, I'll try these methods

Answer 11

Because is such a small number you don't have to use Fermat's little theorem. You can simply say that Mod 3, a can be 0, 1 or 2. Then we have 1^2 = 1 and 2^2 = 4 = 1. So, if a is not zero, a^2 = 1.

Answer 12

convert
36in.=_ft

Answer 13

If a is any intgre

Answer 14

If a is any integer a can be one of the a=3n, a=3n+1 or a=3n+2. In each case you will get 3 as a factor and so...