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Number From Digit Sum (Posted on 2016-05-27) Difficulty: 3 of 5
Find all positive integers that are precisely 2007 times their respective sum of the digits.

*** For an extra challenge, solve this puzzle without using a computer program aided method.

No Solution Yet Submitted by K Sengupta    
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A possible way Comment 1 of 1
If K= Sum of digits of N

N=K*2007 => N = 0 (mod 9) => K= 0 (mod 9) (necessary - not sufficient)


We try: 
K=9    N=K*2007 = 18063 => K=18 (contradictory)
K=18  N=K*2007 = 36126 => K=18 (valid value)
K=27  N=K*2007 = 54189 => K=27 (valid value)
K=36  N=K*2007 = 72252 => K=18 (contradictory)

From here higher values of K (45, 54, 63...) lead to impossibility as a 5 digit N which is multiple of 2007 has his digit sum less than 45; a 6 digit less than 54, so that other matches will be impossible.




  Posted by armando on 2016-05-27 09:20:37
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