Question 657407: A positive integer is divided by 2,3,4, and 5, and the remainders are added together. Find the sum of the digits of the smallest integer that makes the sum as large as possible.
Answer by kevwill(135)   (Show Source):
You can put this solution on YOUR website! In order to maximize the sum of the remainders, we need a number, x, that gives:
(1) remainder 1 when divided by 2, or x = 2m+1
(2) remainder 2 when divided by 3, or x = 3n+2
(3) remainder 3 when divided by 4, or x = 4o+3
(4) remainder 4 when divided by 5, or x = 5p+4
where m, n, o, and p are integers >= 0
From (1) we know the number must be odd, so both n and p must be odd. This gives us
(2a) x = 6n + 5 and
(4a) x = 10p + 9
We can start listing possible values of x for each equation (2a), (3), and (4a):
(2a) 5, 11, 17, 23, 29, 35, 41, 47, 53, 59
(3) 3, 7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59
(4a) 9, 19, 29, 39, 49, 59
59 is the smallest number that meets all the criteria, and the sum of the digits is 14.
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