## Divide number into 3 parts with each part being 1.6 times greater than the last

A friend posed this just for fun but now its really annoying me.

how do you divide 90 into three parts so that each part is 1.6 times greater than the last.
i.e: the second value should be 1.6 times greater than the first and the third value should be 1.6 times greater than the second?

Im confusing myself with this.
Thanks

 Recognitions: Science Advisor Hi CF.Gauss. Just write it as $x + 1.6 x + 1.6^2 x = 90$ and then factorize out the "x" and simplify.

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 Quote by uart Hi CF.Gauss. Just write it as $x + 1.6 x + 1.6^2 x = 90$ and then factorize out the "x" and simplify.
Hm ... I thought the goal here was to help people think and understand things, not spoon-feed them answers. Have I got that wrong?

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## Divide number into 3 parts with each part being 1.6 times greater than the last

 Quote by phinds Hm ... I thought the goal here was to help people think and understand things, not spoon-feed them answers. Have I got that wrong?
Actually I did not give the answer, I left the factorizing and the following arithmetic for the OP to do. This is in effect the first line of what probably be a three line derivation for the OP.

I agree though that it does give away a large part of the overall solution. Sometimes with such a simple question it's hard to know how to give the OP a "start" without giving away too much.

BTW. To me this looked more like a curiosity question than homework anyway, though of course I don't know that for sure.

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 Quote by uart Actually I did not give the answer, I left the factorizing and the following arithmetic for the OP to do. This is in effect the first line of what probably be a three line derivation for the OP. I agree though that it does give away a large part of the overall solution. Sometimes with such a simple question it's hard to know how to give the OP a "start" without giving away too much. BTW. To me this looked more like a curiosity question than homework anyway, though of course I don't know that for sure.
Yeah, I can't argue w/ that. Still, I was going to try to lead him to an equation rather than give it to him.

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