You complete me
× □ 1 0 9 9 1 □ □ □ □ 1 1 □ 0 □ 9 6 9 0 8 0 0 8 7 □ 1 □ □ □ 9 1 1
Above shows an incomplete long multiplication. What is the maximum number of digits I can still remove to allow only one unique solution?
Details and Assumptions :
- For example, I can remove one of the number "1" in the very last column and I'm still able to find the unique solution. So I can remove at least more digit to allow one unique solution.
The answer is 14.
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2 solutions
Moderator note:
Is there a specific reason why you chose two prime numbers? What if they are both composite numbers instead? Can we remove more or less digits?
I think we can remove the 1 in the second line.
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Then 1 9 7 9 × 9 0 9 = 1 7 8 9 1 1 is also another solution which makes the solution not unique.
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This solution doesn't fit the format of problem. We will have a line with five squares instead of four squares.
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@Abdelhamid Saadi – Then the first square has a number 0.
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@Pi Han Goh – What I mean is that for the solution 1 9 7 9 × 9 0 9 the fifth line will be 1 7 8 1 1 which need five squares instead of four squares.
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@Abdelhamid Saadi – Oh, then take 1 9 7 9 × 5 0 9 .
First i would say that you have added 1 extra block in last(answer) line, between 9 and 6..
How do you frame such questions??
Then what is the very last box suppose to be?
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I didn't got you..i meant to say that answer will have 7 digits..and in the problem it seems to have 8 digits..
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Oh, okay edited. But what is the very last box suppose to be?
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@Pi Han Goh – Answer 1996811
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@Charlz Charlizard – No. I could have 1039 * 1103 = 1146017 which fits your long multiplication as well, so your answer is not unique.
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@Pi Han Goh – Yeah..you are correct..i have edited the solution...
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@Charlz Charlizard – Oh okay, your solution looks very similar to mine, you could move the 3 digits in the middle rows to the top and bottom rows to make it similar. But you need to explain why 14 is the maximum.
Let Y and Z denote the values of box in the first row and the box in the second row respectively. Beucase the last digit of 9 Y is 1, then Y has to be 9 only. And if Z is any integer larger than 0, then the product is greater or equals to 1 9 7 9 × 1 0 1 9 > 2 0 0 0 0 0 0 which contradicts the fact the leading digit of their multiplcation is 1.
So the original multiplication is 1 9 7 9 × 1 0 0 9 = 1 9 9 8 6 1 1 .
It's easy to verify that both 1 9 7 9 and 1 0 0 9 are prime numbers so we could not modify the digits in the first two rows. Thus we can remove all the digits in the third to sixth row:
× □ 1 □ □ 9 □ □ □ □ □ 1 1 □ □ □ □ 6 9 0 □ □ □ 8 7 □ □ □ □ □ 9 □ 1
And because we must keep of the last digit and the first digit of the product, we can remove all the other digits the very last row.
× □ 1 □ □ □ □ □ □ □ □ 1 1 □ □ □ □ □ 9 0 □ □ □ □ 7 □ □ □ □ □ 9 □ 1
However, we can still improve on reducing the digits given by noting that 1 9 7 9 × ( 1 0 0 9 + W ) > 2 × 1 0 6 if W > 0 , so we still can remove the zero in the second row.
× □ 1 □ □ □ □ □ □ □ □ 1 1 □ □ □ □ □ 9 □ □ □ □ □ 7 □ □ □ □ □ 9 □ 1
We shouldn't remove any of the 1 's in the first two rows else we can see that there is multiple solutions.
And if we remove any of 9 or 7 in the very first row, there wouldn't be a unique solution, just take 1 9 6 9 × 1 0 0 9 as an example. Hence we have removed the maximum number of digits, which is 14.