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Question

Arrange the numbers $1$ to $9$ to replace letters $A$ to $I$ so:

$(A+B+C+D)-(E+F+G+H) = I$

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score 58 · Answer 1 · answered Jul 11 '20 at 09:38

58

The equation is equivalent to:

$$A + B + C + D = I + E + F + G + H$$

As $A$ to $I$ are $1$ to $9$:

There are $5$ odd numbers, and we couldn't put them on both sides (as the parity will be different!) So it's Impossible.

answered Jul 11 '20 at 09:38

athin

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Much more succinct and elegant than mine, nice! – MrSethward Jul 11 '20 at 09:41
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A slightly different final step: rot13(Gur qvtvgf 1 guebhtu 9 nqq hc gb 45. Vs gur fhz bs nal fhofrg rdhnyf gur fhz bs vgf pbzcyrzrag, gura gung fhz zhfg or unys gur fhz bs gur jubyr frg, juvpu va guvf pnfr vf 22.5, vzcbffvoyr gb npuvrir ol nqqvat bayl vagrtref.) – Joseph Sible-Reinstate Monica Jul 12 '20 at 00:23
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This is really clean – Joe Jul 13 '20 at 09:15
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I suggest this should be the accepted answer. – Dmitry Kamenetsky Jul 13 '20 at 14:53

score 33 · Accepted Answer · answered Jul 11 '20 at 09:37

We know there are 4 evens and 5 odds in the set 1:9.

#1

Let's assume I is odd, then the left side of the equation consists of 4 odds and 4 evens. We can rearrange the equation to separately sum/substract the 4 odds together, which will be even, and 4 evens together, which will be even. Summing those together implies the left hand side of the equation will be even, a contradiction.

#2

Let's assume I is even, then the left side of the equation consists of 5 odds and 3 evens. We can rearrange the equation to separately sum/subtract the 5 odds together, which will be odd, and 3 evens, which will be even. Summing those two together implies the left hand side of the equation will be odd, a contradiction

Thus:

There is no solution to the provided equation!

Oops I just saw your answer as I used mobile >< (perhaps that's why I typed less lol.) But great answer too! :D — athin, Jul 11 '20 at 10:30

score 23 · Answer 3 · answered Jul 13 '20 at 06:52

Thought I'd share my algebraic solution

We know:

(A + B + C + D) - (E + F + G + H) = I

And:

A + B + C + D + E + F + G + H + I = 45

Therefore:

45 - (A + B + C + D + E + F + G + H) = I

45 - (A + B + C + D + E + F + G + H) = (A + B + C + D) - (E + F + G + H)

45 - 2(A + B + C + D) = 0

A + B + C + D = 22.5

Meaning A, B, C, D are not all integers, so there is no solution

score 14 · Answer 4 · edited Jul 13 '20 at 00:09

14

A slightly shorter answer:

Since plus and minus are equivalent modulo 2, your statement implies $$1 = \sum_{i=1}^9 i = A + B + \cdots + H + I = 0 \pmod{2}$$ a contradiction.

edited Jul 13 '20 at 00:09

athin

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answered Jul 12 '20 at 20:57

karu

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score 5 · Answer 5 · answered Jul 14 '20 at 09:59

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We know that 1 + 2 + ... + 9 = 45. We need to solve A + B + C + D = E + F + G + H + I, which means we have to split 45 into two equal parts, which is impossible with integers.

answered Jul 14 '20 at 09:59

Corleone

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So clean! (And so high on the completeness per character scale.) – humn Jul 15 '20 at 01:00
Elegant and easy to understand. – asg Jul 27 '20 at 19:00

score 4 · Answer 6 · edited Jul 13 '20 at 07:59

Adding or subtracting 2 odd/even numbers is even. From 1 - 9 we have 5 odd numbers and 4 even numbers. The above equation

A + B + C + D - (E + F + G + H) = I

Can be rearranged as

A + B + C + D - (E + F + G + H + I) = 0

Adding or subtracting 4 even numbers will result in even number similarly adding or subtracting 4 odd numbers will be even. So we are left with 1 odd number

4 ( odd ) +/- 4 ( even ) +/- 1 odd = 0

Results in

Even +/- Even +/- odd = 0

Which is not possible since even plus odd is always odd.

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6 Answers6