Find all 3-digit integers n (no leading zeroes) such that the sum of the
squares of the digits of n is exactly one-half of n.
For n = 100 To 999
sumsq = 0
ns$ = LTrim(Str(n))
For i = 1 To Len(ns)
sumsq = sumsq + Val(Mid(ns, i, 1)) * Val(Mid(ns, i, 1))
Next
If 2 * sumsq = n Then Text1.Text = Text1.Text & n & crlf
DoEvents
Next
finds
298 is the only such number in the range of 3-digit numbers.
4+81+64 = 149 = 298/2
Outside the range, only 50 fits the bill.
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Posted by Charlie
on 2019-06-05 12:29:13 |
computer solution
