Timeline for Reconstruct a long division given less than a quarter of the digits, and all of those are wrong
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Jun 2 at 15:40 | history | edited | ACB | CC BY-SA 4.0 |
Corrections (credit to @Rosie F)
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| Jun 2 at 15:32 | comment | added | ACB | You are correct. I will edit that part in my answer. Thank you @RosieF. | |
| Jun 2 at 14:39 | comment | added | Rosie F | @ACB Indeed I am. At the start of the solving path, the divisor might be as large as 99. The first subtraction might be, say, 101-51=50 putting a digit 1 in the quotient (if the divisor were 51). It's not until you prove that the divisor is at most 49 that you know that such a situation is impossible. | |
| Jun 2 at 12:59 | comment | added | ACB | @RosieF we are talking about this part, right? I am not seeing the necessity of your argument. | |
| Jun 2 at 12:49 | comment | added | Rosie F | "Consider the first digit of the quotient. It cannot clearly be 1, because the divisor is 2-digit while the dividend is 3-digit." This doesn't follow at this stage. Rather, consider the last digit of the quotient. It is not 1. So 2 or more times the divisor is 2-digit, so the divisor is at most 49. Now consider the first digit of the quotient. It cannot be 1, because, as you say, the dividend is 3-digit. | |
| Jun 1 at 11:40 | comment | added | ACB | Thanks. I'll definitely try it. | |
| Jun 1 at 11:38 | comment | added | user23087 | Well done. Your solution path is pretty much identical to my own. You might be interested in a similar puzzle I just posted. This time one of the digits is correct, but all of the others are wrong. | |
| Jun 1 at 11:07 | vote | accept | user23087 | ||
| Jun 1 at 10:29 | history | answered | ACB | CC BY-SA 4.0 |
