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# Number Theory Base Question

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70
5

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!

Apr 7, 2020

#1
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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!

Apr 7, 2020
#3
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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
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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
#2
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https://web2.0calc.com/questions/number-theory-help_6   Apr 7, 2020
#5
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Yeah, that was correct...thanks!

JustAnotherFailure  Apr 7, 2020