Hmm!
My initial quick guess was 15 (each left hand side number is the difference between the two numbers above it). However, unless the final number 7 is an error (8 according to my logic), that doesn’t fit.
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Hi Steve,
That’s exactly what I thought when I first encountered this teaser 
And IMHO, it would make a good teaser in its own right.
The final number 7 is not a misprint.
It’s just that at present you haven’t managed to visualise the ‘rules’ that will lead to that number 7.
I found it very difficult to ditch my initial ‘logical’ approach, (which was the same as yours and I was really chuffed with it). It took a while before I spotted a different set of rules.
But well-done for getting 15 in the missing square. Let’s say, 4/10 at present ?
d?
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Okay, how about 12?
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Luckily Pinocchio only told 9 lies.
Question to all on this thread: How long would his nose be if he had told 64 lies?
(A variation of the grains of rice on a chess board puzzle)
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12
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Don
12
Hi Steve, Hi Winky,
Spot-on both of you. Well done.
I’ll leave it for a while for either of you to explain the reasoning to the Forum.
Once people see Steve’s first potential (but short of the mark) solution, it’s incredibly difficult for many of them to spot the correct solution. However, when they do … wow - dead easy
Well done Mike, good that you know the size of a tennis court
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I get 12, being the sum of the individual digits of each number on the preceding line.
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Nicely done Mike.
Did you get 15 first ? (and wonder about the 7) Or did you get 12 straight away ?
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I’d read the discussion of the rouge 15 just after I got 15 myself . I hadn’t actually checked the 7, as I thought I had the pattern working down the diagram.
Nothing, nothing, nothing …
4! is 4 x 3 x 2 x 1 = 24
10! is 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800
100! is 100 x 99 x 98 x 97 x … x 2 x 1 = a very big number with lots of zeros at the end of it.
Well, exactly how may zeros does the number 100! have at the end of it ?
PS I found this one to be a toughie. So I won’t be surprised if there is ‘zero’ response
I’ll go with 24.
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@jrhardee’s answer looks good to me. I hope to see the rationale or workings.
I had an unfair advantage on this teaser as a recent project of mine was to see how big an exact factorial I could get on my old 2014 desktop PC. I got to 4,400,000,000! which was almost at the maximum 16GB limit in length (about 41 billion decimal figures).
The direct relevance to this teaser is that I needed to work out exactly how many trailing binary zeroes there were in the answer: 4,399,999,991, which allowed me to save 0.55GB of memory without losing exactness.
Mrs R reckons I should get out more.
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Morning JR. Well done, 24 it is!
I noticed your earlier post in which you were well and truely on the correct path in explaining your method. It’s a difficult one to explain, at least IMHO. I’ll leave it for a day or two in the hope that someone is willing to post their version. My version, similar to yours, involves counting factors of 10 and 5.
I’ll give other people a shot, but I can chime in later if needed.
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Sounds like a good plan.
It might keep one or two forum members busy and therefor out of trouble for the weekend
Not me - I’ve been told to get out more!
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As there are plenty enough factors of 2 (for example there are 50 even numbers), we don’t need to count them. Instead just count the 5’s and combine each with a convenient 2.
There are 20 multiples of 5 up to 100, namely: 5, 10, 15… 95, 100.
There are 4 multiples of 25, namely: 25, 50, 75, 100. Each gives a second factor of 5.
Hence 24 overall. I’m sure there are other approaches too.
Follow up teasers:
How many zeros are there at the end of 1,000! What about 10,000!
What pattern is emerging?