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Aloiza [94]
3 years ago
14

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

Mathematics
1 answer:
fredd [130]3 years ago
6 0

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.

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<h2><u>Value of F </u></h2>

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