Math Brain Teaser: Solve for 100 Using Digits 1-9 | Step-by-Step Guide

[Mk]

How would I hypothetically, go about solving this problem:

Replace each G with one of the digits from 1 through 9 to make a true statement, each digit may only be used once.

G/GxG+GxGxG/G+GxG=100

Thanks guys,
Mk.

 

[arunbg]

How about hypothetical trial and error??

 

[Mk]

I know, but that is so dirty!

 

[NateTG]

Well, you might want to look into what digits are divisible by others, that should reduce the number of possibilities a bit...

 

[hypermonkey2]

Well, if its any help, Richard Feynman once said that there need not be a methodical solution to every problem. If you can tell just by looking at 3x+1=10 that the answer x=3 works, then it is a valid solution. (he said this in reference to a friend who had trouble with his algebra, but knew the answers anyways). Anyways, that just means that any way of getting to the answer is a good one. So don't feel dirty!

 

[topsquark]

Well, if its any help, Richard Feynman once said that there need not be a methodical solution to every problem. If you can tell just by looking at 3x+1=10 that the answer x=3 works, then it is a valid solution. (he said this in reference to a friend who had trouble with his algebra, but knew the answers anyways). Anyways, that just means that any way of getting to the answer is a good one. So don't feel dirty!

Sounds like a Physics professor I had who would solve differential equations by writing the answers down and then verify them. (He'd tinker a bit if he was wrong. But it was fascinating to watch.)

-Dan

 

[hypermonkey2]

Probably was a fan of Feynman (its kinda hard not to be these days). In any case, I would help find the solution, except it reminds me too much of Su Doku. Best of luck!

 

[arunbg]

The Answer

The answer to the question is given below in white

6/(2*1) + 8*7*5/4 + 9*3 =100

Q: How did I do it?
A: Hypothetical trial and error (at least that is what my computer did)