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The sum of three consecutive multiples of 9 is 567. Find the largest of these numbers.

User Hscasn
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Answer:

198

Explanation:

Interesting question. Not sure if there would be an equation for it, but we can think through it pretty easily.

First let's make sure we understand what "multiples of 9" means:

9, 18, 27, 36, 45,.... Agree?

Then we have the word "consecutive," and that essentially means next to each other, right? So (27, 36, 45) would be "three consecutive multiples of 9." Good so far?

Okay, here's where it gets jus a tiny bit tricky. Ooooo, and I just thought of a formula for it! (Which I'll lay out after this thought experiment.)

Since the three numbers will be about the same size, then each will be about one-third of 567. Do you see that?

So 567 ÷ 3 = 189

Is that divisible by 9? Yes, 21 times, so it's a multiple of 9. Hmmmm, maybe that's the middle one, and the others are one multiple below (180) and one multiple above (198). Let's see if those work:

180 + 189 + 198 = 567

Yay, that worked out, and 198 is the answer.

But to do it algebraically, let's think of it like this:

We need a number, x, that if we add to it two other numbers that differ from it by only 9, those three added up would equal 567. How about:

(x - 9) + x + (x + 9) = 567

Do you see how that would work? It's three consecutive numbers, one of them a multiple of 9 lower, and one a multiple higher, and those terms must add up to 567.

So let's work that out:

x - 9 + x + x + 9 = 567

3x = 567

x = 567/3 = 189

See how the 9's cancelled out? And more importantly, how we ended up at x = 189? That's what we'd come up with just by thinking about it, remember?

But is 189 the answer? No, it's the middle number. To get the "largest" number we just add 9 of course, to get 198, same as before.

If you want to test your understanding of this concept, why does this equation also work:

(x - 18) + (x - 9) + x = 567

It does work, and in this case solving for x gives you the highest number directly.

User Dinkheller
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