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If n is natural number , then 6n2+6n is always divisible b

If n is a natural number, then

(6n2 6n) is always divisible

(6n2 6n) is always divisible

If n is a natural number, then

(6n2 + 6n) is always divisibl

(6n2 + 6n) is always divisibl

Let A be a natural number consisting only 1. B is another na

Which one is correct statement? Circuit is always openNetwor

If we write down all the natural numbers from 259 to 492 sid

If we write down all the natural numbers from 259 to 492 sid

It is given that 232 1 is exactly divisible by a certain num

If

(232+1) is completely divisible by a whole number, which

(232+1) is completely divisible by a whole number, which

It is being given that

(232 +

1) is completely divisible by

(232 +

1) is completely divisible by