Consider your calendar, displaying a month.
Select three consecutive dates in a row, column or diagonal.
Write them one beside another to form a single number.
Call it A, the 1st number.
In a similar way select another three dates and write them down to get B,
the 2nd number. Multiply A by B.
Give me the list of digits of the product skipping just
one non-zero digit.
I'll immediately disclose the digit you have skipped.
Why does it work?
Three numbers in a row will total 3 times the middle number. Of course you are not adding them strictly speaking, but just as the sum is a multiple of 3, so is the sum of the digits, and therefore the number formed by concatenation is a multiple of 3.
The same is true of diagonals n-8 + n + n+8 = 3*n, and the sum of digits again is a multiple of three and therefore also the contatenation. And the same with the other diagonal n-6 + n + n+6, or columns n-7 + n + n+7.
Multiply them together you get a multiple of 9. The sum of digits must be a multiple of 9. You just quote the single-digit number needed to bring the total to a multiple of 9.
6 3
13 and 11
20 19
61320 is a multiple of 3
31119 is a multiple of 3
61320 * 31119 = 1908217080, a multiple of 9 and indeed its digits add up to 36, a multiple of 9. Whatever digit is left out will result in a number that's not a multiple of 9, and you add what's needed to get to the next multiple of 9, i.e., to get to 36.
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Posted by Charlie
on 2016-09-09 10:54:07 |
solution
