Only 1% of people pass this logic test

This is one of the most popular puzzles, and there is an incredible way to solve it!

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[gnericc_mistyck]

The only reason why I knew something was up was because 9 should be a stripe, not a solid.

[briant]

Use the 15 twice and turn any of the solids around so the number doesn't show!

[louisauffret]

they're all odd numbers we cant


edit : that's not what I signed up for

[mf]

When he said you can use a ball twice, I immediately thought to just take 15 twice and turn one of the other balls around so the number doesn't show anymore. 15+15+0=30

[nchoosekmath]

Not trying to be that guy, but there should be an underline to indicate whether it's a 6 or a 9.

[eventhisidistaken]

The thing is, once you abandon the apparent restrictions, there are probably an infinite number of solutions. Smash two of them and spray paint one with "30" on it. There, that's just as valid of a "solution".

[alexminsky]

Sorry, but the solution seems more like a cheap trick rather than a full-fledged mathematical method.

[derwolf]

You also could solve that with two 6 balls and a 15 ball:
9+6+15=30.

[SuzakuDragon]

Tired: turning the "9" ball into a 6
Wired: Base 9 arithmetic (13 + 15 + 1 = 30)

[waiyanoo]

Generally on billiard balls, there is a bar under the numbers 6 and 9 to further avoid confusion. Also, to those who might not play the game, the color of 8 is black and 8+n is the same as n, with the bigger number as stripes and the smaller as solids. So green solids is 6 and green stripes is 14. 9 will be a striped yellow. So in theory, the numbers given are actually 1 3 5 7 6 11 13 and 15. Using 9 is more like cheating than using 6.

[MuNu_XiPi]

I believe that more than 1% can solve this.

[Lord_Skeptic]

15+13+1=30 (base 9).
Nowhere does it say it has to be in decimal does it.

[goseigentwitch]

I used kinda of a different cheat. You said that three numbers must be used, but never said three circles must be sued.
I put 1 next to 5 to concatenate into 15 in the 1st circle and a 15 in the 2nd circle.
I used three numbers in only 2 circles but got 15 + 15 = 30

[flexable]

Except that the 6 and 9 balls have a mark to tell the players this is the 9 or the 6.

[LogicalMath]

6 + 9 + 15 = 30 is another solution, as it is allowed to use numbers more than ones (6 and 9 on the same ball)

[robertharrison]

"You're allowed to use a number more than once but you must use exactly three of the balls" I interpreted that to allow me to use the 1, the 13, and the 15 (exactly three of the balls) and I use the 1 more than once (twice), so I get the total of 30.

[kazebaret]

I actually solved the problem assuming that 30 was a number in a base > 3. You can show then that it has to be odd. So it means that also the balls are written in that base. Let's start with base 5. (30)_5 = 15 in base 10. In this case, considering that in base 5 the only digits that exist are 0-4, we can only use the balls 1, 3, 11, 13. (1+13+11)_5 = (3+11+11)_5 = (30)_5. In general, the numbers in the balls can be written as 1, 3, 5, 7, 9, k+1, k+3, k+5. And the sum of 3 of them has to be equal to 3k ((30)_k = 3k), with k odd >3, and you can only use numbers whose digits are < k. Thus, you can show that the maximum base that satisfies the equality is 19: (k+5)+(k+5)+9 = 3k => k =19.

[SteveSmith-rtwx]

Also, depending on how the question is written/presented, you could use more than three balls. I missed the "exactly" in Presh's explanation, so if the presenter forgot that key word then the rules could be taken as: "You can use a ball more than once, but you must use [at least] three balls"

If the answer can be, "Well the nine is actually a six" Then my BS cheat is also legit

[prashantsubedi]

Who remembers this question from the movie ‘Genius‘ ?

[xstuff]

I thought the solution was 11 + 13 + 15: the question was worded as “the numbers on the balls add up the 30”, so you’re not necessarily limited to the two additions in the question, therefore the solution was 1 + 1 + 13 + 15