 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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) Comments: ( Back to comment list | You must be logged in to post comments.) computer solution | Comment 2 of 7 | We need not check farther than 3-digit numbers, as 4 * 9^2 = 324 -- only a 3-digit number; and in fact, 3 * 9^2 = 243, so even that number need not be exceeded.

DECLARE FUNCTION sumsq# (x#)
DEFDBL A-Z
FOR n = 1 TO 243
IF n = sumsq(n) + 1 THEN PRINT n
NEXT n

FUNCTION sumsq (x)
s\$ = LTRIM\$(STR\$(x))
t = 0
FOR i = 1 TO LEN(s\$)
t = t + VAL(MID\$(s\$, i, 1)) * VAL(MID\$(s\$, i, 1))
NEXT
sumsq = t
END FUNCTION

the results being only 35 and 75.

 Posted by Charlie on 2011-09-03 14:56:00 Please log in:
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