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Number = Sum (Digit)*Product (Digit) (Posted on 2010-06-21) Difficulty: 2 of 5
N is a positive base ten integer having at least 2-digits but at most 4-digits, which is obtained by multiplying the sum of its digits with the product of its digits. It is known that N cannot contain any leading zero.

Determine all possible value(s) of N.

No Solution Yet Submitted by K Sengupta    
Rating: 3.0000 (1 votes)

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re: Solution & thoughts | Comment 7 of 10 |
(In reply to Solution & thoughts by Jer)

I have a start to an analytical solution, too tired to go any further today, perhaps somebody can carry on where I have left off.


Now let s be the sum of the digits of n and p be the product of the digits of n, then we have
n=s*p
we also have that 
n mod 9 = s mod 9
so let r = n mod 9 then we have
n=9a+r and s=9b+r

so then we get
9a+r = (9b+r)p
9a+r=9bp+pr
thus we need that
pr mod 9 = r
now pr mod 9 = [r * (p mod 9) ] mod 9
so let p mod 9 = s
then we need
r*s mod 9 = r

now r,s are both on the interval [0,8] and thus we need to find all r,s such that r*s mod 9 = r these are
(r,s): (0,s), (r,1), (3,4), (3,7), (6,4), (6,7)

so what I am thinking can happen is that we can look at each of these possible values for r,s and determine what solutions are possible for n.

Edited on June 21, 2010, 10:58 pm
  Posted by Daniel on 2010-06-21 21:11:59

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