If S, H, and E are all distinct non-zero digits less than 5 and the following is true, find the sum of the three values S, H, and E, expressing your answer in base 5. (There should be a line between the second and thrid columns. Sorry about the LaTeX!)

\(\begin{array}{c@{}c@{}c@{}c} &S&H&E_5\\ &+&H&E_5\\ \ &S&E&S_5\\ \end{array}\)

All help=gratitude. Perferably ASAP, and thank you!

JustAnotherFailure Apr 7, 2020

#1**0 **

The question is actually easy if you think about it. S, H, and E have to be less than 5 since it's in base 5. If 2E=S, S equals either 2 or 4 and since 2H=E, that means that H is 1, E is 2, and S is 4. This is the only case that works. So your answer is 1+2+4= **7**.

Hope it helps!

HELPMEEEEEEEEEEEEE Apr 7, 2020

#3**+1 **

Uh, I don't know why, but it said your answer was wrong. Here's the explanation they gave me:

Starting with the the rightmost column would be the easiest (we'll call it the first, the second rightmost column the second, and so on). Let's consider the possible values for E first.

Since the values must be non-zero, we'll start with 1. If E is 1, then S would be 1 and nothing would carry over. However, since H+H must equal E if nothing carries over and H must be an integer, E can not equal 1.

If E equals 2, then S must equal 4. H would then equal 1. This satisfies our original equation as shown.

I don't know what you did wrong, but hope that helps! Have a nice day :D

JustAnotherFailure
Apr 7, 2020

#4**+1 **

My answer is the same as what you said, I'm confused...

And I quote "that means that H is 1, E is 2, and S is 4."

Both CPhill and I have the same answer... I don't understand??

Wait...I understand, I found the values of E, H, and S but it wanted the sum in base 5 from the diagram. That makes sense.

HELPMEEEEEEEEEEEEE
Apr 7, 2020

edited by
HELPMEEEEEEEEEEEEE
Apr 7, 2020

edited by HELPMEEEEEEEEEEEEE Apr 7, 2020

edited by HELPMEEEEEEEEEEEEE Apr 7, 2020

edited by HELPMEEEEEEEEEEEEE Apr 7, 2020

edited by HELPMEEEEEEEEEEEEE Apr 7, 2020