Consider the 800-digit integer 234523452345......2345.
The first m digits and the last n digits are crossed out so that the sum of the remaining digits is 2345.
Determine the value of m+n
The sod(2345) = 14
The sod of the entire number is 200*14 = 2800.
The sum of the removed digits is 2800 - 2345 = 455.
The removed digits will be 32 groups of 2345 (with sod 448) and one additional 2 and one additional 5.
So (m+n), the number of removed digits, is 32*4 + 2 = 130
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Posted by Larry
on 2023-10-15 12:44:37 |
solution
