if a^2 is divisible by 7 then a is divisble by 7 to?

Answer 1

a^2 = a*a

If a is divisible by 7, then a^2 is also divisible by 7.

Answer 2

Yes. If a were not divisible by 7 then we would have a=7n + b for some nEN, b = 1, 2, 3, 4, 5 or 6.

Then a² = 49n² + 14nb + b² = 7(7n²+2nb) + b², which would mean b² is a multiple of 7. But none of 1, 2, 3, 4, 5, 6 when squared gives a multiple of 7. So this can't happen.

Answer 3

If you are talking only about integers then a would be divisible by 7

a^2 =7 then a =squareroot(7)
a^2 =14 then a =squareroot(7)*squareroot(2)
a^2 =21 then a =squareroot(7)*squareroot(3)
a^2 =35 then a =squareroot(7)*2
a^2 =49 then a =7

a would be divisible by 7 if a^2 is of the form 49*(n^2) in which n is an integer.

Answer 4

Yep! The base has to be a multiple of 7 because it's only being squared. Since 7 is a prime, no other number squared will be divisible by 7.

Answer 5

Yes.

Basic principle of mathematics, called the fundamental theorem of arithmetic. Every number has a factorization and it is unique up to order. So if a is the product of a whole bunch of primes p rasied to whole bunch of powers, then a^2 is the product of the same primes but raised to powers that are twice as large. If the prime 7 is on the list for a^2, then 7 had to be on the list of primes for a as well.

Answer 6

Yes,'a' will also be divisible by 7.

Answer 7

edit
it is true only for prime nos yes and not for composite nos.
let a be 4
a^2=16
a^2 is divisible by 16
but 4 is not
so fallacious statement for non prime nos

Answer 8

If we assume that a is an integer, this is true. However, if a is permitted to be irrational, then you have the obvious counterexample a=√7

Answer 9

yes

Answer 10

Has to be.