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 Squaring Up The Digits II (Posted on 2011-09-03)
Determine all possible values of a base ten positive integer N such that N is precisely one more than the sum of the squares of its digits.

 No Solution Yet Submitted by K Sengupta Rating: 4.3333 (3 votes)

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 Fully explained solution | Comment 5 of 7 |

I've promised a solution.  Here it  goes:

1. It is easy to show that only 2-digit numbers are eligible.
2. Assume the number  is  10*a+b
Then  a^2+b^2+1 = 10*a+b
a*(10-a)-1=b*(b-1)
3. From the above equation we can state:
-If a   is a solution ,so  is  10-a (symmetry)
-a is odd, since b*(b-1) is even
4. Now we solve an equation  b^2-b-k=0,
where k=a*(10-a)-1:

b=1/2*(1+sqrt(1+4*k)) ,   trying k=8,  20,  or  25
(a=1 or 3 or 5)
5. k=20 triggers b=5, a=3 and 10-a=7

)

Edited on September 4, 2011, 4:00 am
 Posted by Ady TZIDON on 2011-09-04 03:54:33

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