Here is a riddle written on a cup:
Eh is four times as much as Oi,
Oh is four times as little as Ai,
What do you get if you add all four of them up?
Source: Russian Olympiad Problems, Math Circle (Beginner) 2018 PDF Q6
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$\begingroup$ Welcome to Puzzling, seems a nice question! $\endgroup$– TomApr 28, 2020 at 15:35
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2$\begingroup$ Old Macdonald ? $\endgroup$– Rand al'ThorApr 28, 2020 at 15:42
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1$\begingroup$ Appears to be from the 2018 russian olympiad (Q6) $\endgroup$– Beastly GerbilApr 28, 2020 at 15:59
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$\begingroup$ This surely seems too broad, unless the cup part is important somehow... but I'm of the opinion it's just for rhyming purposes $\endgroup$– QuintecApr 28, 2020 at 16:55
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$\begingroup$ @BeastlyGerbil Oh I didn't know that! I was told it by a friend. Good find. $\endgroup$– SimdApr 28, 2020 at 17:24
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3 Answers
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Writing equations from the poem:
$EH = 4 \times OI$
$AI = 4 \times OH$
So we know that O is
1 or 2
And then we have
$H \times 4 = xI$
$I \times 4 = xH$
So H and I are
(2,8), (4,6), (6,4), or (8,2)
So OH is
12, 14, 16, 18, 24, 26, or 28
and AI is
48, 56, ~64~, 72, 96, ~104~, or ~112~ (~crossing out~ the ones that duplicate digits or are three digits)
So (OH,AI,OI) are
(12,48,18), (14,56,16), (18,72,12), or (24,96,26)
Then (OH,AI,OI,EH) are
(12,48,18,72), ~(14,56,16,64)~, (18,72,12,48), or ~(24,96,26,104)~ again ~crossing out~ the ones that duplicate digits or are three digits
Which leaves
(12,48,18,72) or (18,72,12,48)
So the sum is
150 no matter which of these you choose
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$\begingroup$ I don't understand the first line of your solution. Why can't O be 0? $\endgroup$– SimdApr 28, 2020 at 21:32
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1$\begingroup$ We don't typically write numbers starting with 0, but I guess you could create a solution where O is 0 if you wanted to. $\endgroup$– shooverApr 28, 2020 at 22:17
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I and H are both even, and, from the rhyme, I+H=10 is easy to deduce. Also O=1 or O=2. But if O=2, then one of OH, OI is greater than $25$. so O=1. The riddle asks for EH+OI+OH+AI=5(OI+OH)=150.
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$\begingroup$ @Anush; $4i\equiv h \pmod{10}$ and $4h\equiv i \pmod{10}$ gives you $h,i\equiv 0 \pmod2$. Adding gives $3(i+h)\equiv 0 \pmod{10} \implies i+h\equiv 0 \pmod{10}$. $\endgroup$– JMPMay 1, 2020 at 8:34
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We have Eh = 4Oi, Ai = 4Oh, which constrains the smaller values to the range 10..25.
So i = 16h, mod 10, which has 2 solutions 0 (reject so that i $\neq$ h), and i = 6.
With the above constraint on Oi, O=1, so Oi = 16, Eh = 64, Oh = 14, and Ai = 56.
The sum of the 4 2-digit numbers is then 150.
