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| Squaring Up The Digits II (Posted on 2011-09-03) |
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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.
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No Solution Yet
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Submitted by K Sengupta
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Rating: 4.3333 (3 votes)
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Possible soution
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 | Comment 4 of 7 | 
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a^2+b^2+1=10a+b 4(a-5)^2+(2b-1)^2 = 97 say, (2x)^2+y^2=97 x=+/-2, y=+/-9 a-5=2,a-5=-2 a={7,3} 2b-1=9, (2b-1=-9) b={5} N={35,75}
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Posted by broll
on 2011-09-04 01:29:59 |
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