(16^3 + 17^3+ 18^3+ 19^3), divided by 70 leaves a remainder ?

(16^3 + 17^3+ 18^3+ 19^3), divided by 70 leaves a remainder ?


There's no remainder, and here's how you can prove it without using a
calculator *or* even bothering to work out all the numbers. It depends
on knowing this key factoring pattern, commonly taught in algebra
classes: for any pair of numbers a and b,

a³ + b³ = (a + b)(a² - ab + b²)

So we start by recognizing that 70 = 35·2. Next, we rearrange the
terms in the problem to look like this:

16³ + 19³ + 17³ + 18³

We can group the terms in pairs like this:

(16³ + 19³) + (17³ + 18³)

Now we factor each pair, using the pattern above:

(16 + 19)(16² - 16·19 + 19²) + (17 + 18)(17² - 17·18 + 18²)

now, 16 + 19 and 17 + 18 both equal 35, so:

(35)(16² - 16·19 + 19²) + (35)(17² - 17·18 + 18²)

We can now factor out the 35:

(35)(16² - 16·19 + 19² + 17² - 17·18 + 18²)

Now, let's look at that long expression in the parentheses. It's the
sum of four even numbers (16², 16·19, 17·18, and 18²) and two odd
numbers (19² and 17²). Since any two odd numbers always add up to an
even number, the terms in the parentheses all add up to some even
number. We don't know what it is, but we don't care -- all we care is
that it's even, which means it contains a factor of 2.

So our number has a factor of 2 and a factor of 35 -- and since 70 =
35·2, our number has a factor of 70. Which means that dividing it by
70 leaves no remainder.

I'm betting your teacher will give more credit for an answer like that
for one like "the calculator says so." :-)

Hope that helps!

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