Prove that 17^16+17^14 is divisible by 5

We'll use the exponent rule that a^(b+c) = a^b*a^c

to break up 17^16 into 17^(14+2) = 17^14*17^2

That allows us to factor like so

17^16+17^14

17^14*17^2+17^14*1

17^14(17^2 + 1)

Computing 17^14 is a lot more difficult compared to computing 17^2 + 1, assuming you only could use paper and pencil.

Note that 17^2 = 17*17 = 289, so 17^2 + 1 = 289+1 = 290

Because 290 ends in a 0, this means that 5 is a factor

290 = 5*58

This makes 5 to be a factor of the original expression.

So overall,

17^16+17^14 = 17^14(17^2 + 1) = 17^14*290 = 17^14*5*58 = 5*17^14*58

The stuff "17^14*58" doesn't matter. All we care about is that factor of 5.

Since 5 is a factor, this means that 17^16+17^14 is a multiple of 5, or 17^16+17^14 is divisible by 5.

Phrased another way, 17^16+17^14 all divided over 5 leads to some integer. In this case, the integer would be the result of 17^14*58.

step-by-step explanation:

subtract 7, so the answer is x < = 3, or c.

the sum of all the angles in a triangle are equal to 180

so you can add   the 3 given angles together to equal 180:

5y + 2y + 2y = 180

or since all three angles have the same letter, combine the 3 angles to equal 180:

9y = 180