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Christine
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15-digit number
« on: Mar 23rd, 2013, 12:46pm » |
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This is a multiplication puzzle. The goal is to find the largest product.
Take a zeroless 15-digit integer, e.g. 283449762837692
Then, insert 4 multiplication signs inside the number sequence and obtain a multiplication of 5 numbers.
Find the largest product. Describe your method.
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towr
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Re: 15-digit number
« Reply #1 on: Mar 23rd, 2013, 2:50pm » |
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A good start might be to put a multiplication in front of the four largest digits (ignoring the first).
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pex
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Re: 15-digit number
« Reply #2 on: Mar 23rd, 2013, 3:46pm » |
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on Mar 23rd, 2013, 2:50pm, towr wrote:| A good start might be to put a multiplication in front of the four largest digits (ignoring the first). |
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That's a good start, but we'd still need some sort of non-obvious (to me) heuristic to lead us to the optimal solution. In the given example, it would yield
2 x 8344 x 9762 x 8376 x 92 = 125535798807552, while
28344 x 9762 x 83 x 76 x 92 = 160575563467008 is almost 28% more.
on Mar 23rd, 2013, 12:46pm, Christine wrote:I'd be surprised if anything but brute force could be guaranteed to work.
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towr
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Re: 15-digit number
« Reply #3 on: Mar 24th, 2013, 8:16am » |
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Fortunately, brute-force consists of trying only 3876 options.
On the other hand, it may not be so hard to spot that 2.8/2 > 8/7; I think it's mostly the first number that's a real problem.
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I wonder what happens if you keep repeating the step; the number should get smaller each step, so eventually you should get a single digit.
(Of course we'd have to decide first what to do when we have less than 15 digits; once we have less than 5 it doesn't make sense to try and put 4 multiplications in. How about ceil(#digits/3 -1) multiplication signs for each step? --- Which would mean we stop at a 3 digit number or less.)
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pex
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Re: 15-digit number
« Reply #4 on: Mar 24th, 2013, 8:45am » |
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on Mar 24th, 2013, 8:16am, towr wrote:| Fortunately, brute-force consists of trying only 3876 options. |
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Even fewer: there are only 14 spots for the multiplication signs, not 19  , which makes it 1001 options.
on Mar 24th, 2013, 8:16am, towr wrote:I wonder what happens if you keep repeating the step; the number should get smaller each step, so eventually you should get a single digit.
(Of course we'd have to decide first what to do when we have less than 15 digits; once we have less than 5 it doesn't make sense to try and put 4 multiplications in. How about ceil(#digits/3 -1) multiplication signs for each step? --- Which would mean we stop at a 3 digit number or less.) |
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Interesting extension! We do need to worry about trailing zeros, however...
Using the same example as before:
28344 x 9762 x 83 x 76 x 92 = 160575563467008
1605 x 755 x 6346 x 700 x 8 = 43063575240000
430 x 63 x 5 x 7 x 5240000 = 4968306000000
4 x 9 x 6 x 830 x 6000000 = 1075680000000
10 x 7 x 5 x 6 x 80000000 = 168000000000
and any attempt to insert four multiplication signs will make the result zero.
What if we cut off trailing zeros?
28344 x 9762 x 83 x 76 x 92 = 160575563467008
1605 x 755 x 6346 x 700 x 8 = 43063575240000, trimmed to 4306357524
430 x 63 x 5 x 7524 = 1019125800, trimmed to 10191258
101 x 9125 x 8 = 7373000, trimmed to 7373
73 x 73 = 5329
532 x 9 = 4788
47 x 88 = 4136
413 x 6 = 2478
247 x 8 = 1976
19 x 76 = 1444
14 x 44 = 616
and we're done.
However, if we had used the information that trailing zeros would be cut off, we could have made different decisions along the way to make the resulting product larger... hmmm...
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ravibhole_1
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Re: 15-digit number
« Reply #5 on: Apr 7th, 2013, 12:20am » |
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Take a zeroless 15-digit integer, e.g. 283449762837692 (may be repeated digits
9999x9999x9999x9999x999
= 99980001x99980001x999
This will be largest product
OR
9876x9876x9876x9876x987
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