Arrange the numbers $1$ to $9$ to replace letters $A$ to $I$ so:
$(A+B+C+D)-(E+F+G+H) = I$
Too easy? Too many answers? Try it first! Then explain why.
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asked Jul 11, 2020 at 9:12
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6 Answers
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1
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!
answered Jul 11, 2020 at 9:37
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1$\begingroup$ Oops I just saw your answer as I used mobile >< (perhaps that's why I typed less lol.) But great answer too! :D $\endgroup$– athinJul 11, 2020 at 10:30
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4
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.
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5$\begingroup$ Much more succinct and elegant than mine, nice! $\endgroup$ Jul 11, 2020 at 9:41
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4$\begingroup$ 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.) $\endgroup$ Jul 12, 2020 at 0:23
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1$\begingroup$ I suggest this should be the accepted answer. $\endgroup$ Jul 13, 2020 at 14:53
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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
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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.
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2
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.
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$\begingroup$ So clean! (And so high on the completeness per character scale.) $\endgroup$– humnJul 15, 2020 at 1:00
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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.
