This is not homework – this is from a sheet of sample questions from an IQ test that I was given. There were 6 questions. The first four were ridiculously easy. The fifth was a bit harder, but not too bad after a few minutes of thought. But the 6th? I can’t figure it out. I’m sure I’m missing something obvious, and one of you will get it in seconds.
There’s a visual component, so I took a photo and uploaded it to imgur.
As they used to say in school, plese explain your answer.
Thanks!
11
Add up the digits
2+5+5= 12, 3+6+5= 14, 4+9+5= 18, 1+5+5= ?11?
That’s it! Thanks!
You’re welcome, I’m pretty surprised this didn’t gather more interest. I love a good brain teaser.
That’s a good one. Here is a visual one - get ready to draw:
Draw two squares on top of, and centered on, three squares. All squares the same size. Looks like a pyramid, minus the top square (three on bottom, two on top).
OK, now, with a pencil, draw one continuous line, without lifting-up your pencil, that bisects each and every line segment (including the half-segments between the two layers. You can start anywhere, except on a line. Note, there are 14 line segments.
That’s it. Good luck!
Like this?
…±–±–+
…|…|…|
…|…|…|
±±±±±±+
|…|…|…|
|…|…|…|
±–±–±–+
If so, I count 18 edges that need to be bisected – 2 horizontal on top, 3 vertical on the top row of squares, 6 short horizontal in the middle, 4 vertical on the bottom row, and 3 horizontal on the bottom. If this is what you meant…
Is it a trick question, where you’re actually allowed to cross/touch the line you made or cross an edge more than once? That makes it pretty trivial. Otherwise, I’m like 95% sure it can’t be solved. Since the bottom left square has 5 edges to be bisected, the line must start or end inside of it. This reasoning applies to the other 4 squares as well – they all have 5 edges needing to be bisected. So, unless the line has 5 ends, the problem can’t be solved.
I think this is wrong - it should be 64. On one side you have a simple square, on the other the digits of the square plus 5 added up. since 15 is not a square, it must be the digits added up; the opposite has to be 6 + 4 (+5) = 15.
I’m certain that can’t be done. I’ve known about it since the early 80’s and have never seen it solved.
If that’s the pattern, wouldn’t 19, 28, 37, etc. also be acceptable solutions?
It’s not like there are any directional markings either way on the lines…
I think 11 is more likely because there are multiple indicators of directionality:
If you go by the squares rule, you violate the other rules.
Also because doing it your way would give you too many possible answers. 15-5 = 10, which could be 91, 19, 64, 37, etc.
They only go one way, not both ways… and given the above, it seems more likely that 15 is the origin.
Kimble, yes, that is the correct configuration. And, yes, I neglected to add the rule that the drawn line can only bisect each segment once.
G0sp3l, that is my assessment as well. I was thinking someone here may have seen a solution for it.
OK, back to the OP.
Proof that the five-squares one is imposslble:
[spoiler]Place points A and B in the squares on the top row, points C, D, and E on the squares in the bottom row, and point F anywhere outside any of the squares.
If you connect the points with one line through each line segment, you get 5 lines going into C (one from A, one from D, and three from F, through the top, left, and bottom of square C), 5 lines going into D (one from each other square, and one from F through the bottom of square D), and 5 lines going into E (one from B, one from D, and three from F, through the top, right, and bottom of square E).
In order to have a path, other than the starting and ending points, the number of paths into a point must equal the number of paths out of a point. This means that no more than 2 points can have an odd number of paths, but C, D, and E each have 5.
[/spoiler]
This is the answer I came to, as well. I think I like the 11 one a little better, but they’re both acceptable, I should think.
I came up with 11 originally, based on the digits adding to the bottom number. I think 64 is the answer they were going for, though.
Both answers are justifiable, though, and the puzzle should have been better designed to avoid the ambiguity.
I still can’t see how 11 is justifiable; there is no consistent pattern that holds true for every number.
Nebver mind, 11 works too.