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
Puzzling Stack Exchange is a question and answer site for those who create, solve, and study puzzles. It only takes a minute to sign up.Sign up to join this community
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
(12,48,18,72) or (18,72,12,48)
So the sum is
150 no matter which of these you choose
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.