5
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Welcome to Puzzling, seems a nice question! $\endgroup$ – Tom Apr 28 at 15:35
  • 2
    $\begingroup$ Old Macdonald ? $\endgroup$ – Rand al'Thor Apr 28 at 15:42
  • 1
    $\begingroup$ Appears to be from the 2018 russian olympiad (Q6) $\endgroup$ – Beastly Gerbil Apr 28 at 15:59
  • $\begingroup$ This surely seems too broad, unless the cup part is important somehow... but I'm of the opinion it's just for rhyming purposes $\endgroup$ – Quintec Apr 28 at 16:55
  • $\begingroup$ @BeastlyGerbil Oh I didn't know that! I was told it by a friend. Good find. $\endgroup$ – fomin Apr 28 at 17:24
7
$\begingroup$

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

| improve this answer | |
$\endgroup$
  • $\begingroup$ I don't understand the first line of your solution. Why can't O be 0? $\endgroup$ – fomin Apr 28 at 21:32
  • 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. $\endgroup$ – shoover Apr 28 at 22:17
4
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ How are you using the rhyme exactly? $\endgroup$ – Anush May 1 at 8:24
  • $\begingroup$ @Anush; $4i\equiv h \pmod{10}$ and $4h\equiv i \pmod{10}$ gives you $h,i\equiv 0 \pmod2$. Adding gives $3(i+h)\equiv 0 \pmod{10} \implies i+h\equiv 0 \pmod{10}$. $\endgroup$ – JMP May 1 at 8:34
1
$\begingroup$

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.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.