Which of the following can be used to illustrate that not all prime numbers are odd? 

 

Given a x a = 12

asked to find a x a x a x a =  (a x a) x (a x a)

a⁴ = a² x a² 

= 12 x 12

= 144.

More :: Mathematical operations questions and answers.

1 < 4n + 7 < 200

n can be 0, or -1

n cannot be -2 or any other negative integer or the expression 4n + 7 will be less than1.

The largest value for n will be an integer < (200 - 7) /4

193/4 = 48.25, hence 48

The number of integers between -1 and 48 inclusive is 50

The fence will consist of one more post than there are chains. (e.g. P-c-P-c-P).

Therefore, a total length has to be a multiple of the length of the chain plus one post (5.5) plus one post extra.We have length = (5.5n + 0.5), where n can be any positive whole number. If n= 3, length =17

Let the required fraction be x. Then, (1 / x )- x = 9/20
1 - x^(2) / x = 9 / 20  =>  20 - 20 * x^(2) = 9 * x.
20 * x^(2) + 9 *x - 20 = 0.
=> (4 * x + 5) (5 * x - 4) = 0.
=> x = 4 / 5.

6n2+6n=6nn+1, which is always divisible by 6 and 12 both, since n(n+1) is always even.

987 = 3 * 7 * 47.
So, the required number must be divisible by each one of 3, 7, 47
553681 => (Sum of digits = 28, not divisible by 3)
555181 => (Sum of digits = 25, not divisible by 3)
555681 is divisible by each one of 3, 7, 47.

Here a = 3 and r = 6/3 = 2. Let the number of terms be n.
Then, t = 384 => a * r^(n-1) = 384
=> 3 * 2^(n-1) = 384 => 2^(n-1) = 128 = 2^(7)

=> n-1 = 7 => n = 8.

6)4456(742
42
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25
24
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16
12
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4
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Required number = (6-4) = 2.