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A Math Problem (Posted on 2015-12-03) Difficulty: 3 of 5

Alan, a student teacher, was asking for advice in the math common room.

'I started with Perplexus puzzle 3185,' he explained:-

'There are pairs of numbers whose sum and product are perfect squares. For instance, 5 + 20 = 25 and 5 x 20 = 100. If the smallest number of such a pair is 1090, what is the smallest possible value of the other number?'

'To make it more topical, I wanted to replace the smallest number with 2015, but I can't seem to find a solution.'

What if you wait until New Year, and use 2016 instead?' suggested Ms. Brown, a lecturer.

Professor Croak chipped in. 'That still would not work, as the problem stands. One alternative would be to wait until 2017, but the solution might be very difficult! Another would be to do as Ms. Brown suggested, but with a minor change to the wording of the problem, leading to a neat result that ought to please the most demanding pupil.'

What change did Professor Croak propose, and why?

See The Solution Submitted by broll    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution computer solution of a way that works | Comment 1 of 6
Instead of "sum", use "absolute difference".

When this is done, with 2016 being the lower number, the higher number is 453600. The difference and product are 451584 and 914457600 respectively, which are 672^2 and 30240^2 respectively.

DefDbl A-Z
Dim crlf$

Private Sub Form_Load()
 Form1.Visible = True
 Text1.Text = ""
 crlf = Chr$(13) + Chr$(10)
 For tot = 4033 To 1000000
     a = 2016
     b = tot - a
     If b <> a Then
       diff = b - a
       prod = b * a
       sr1 = Int(Sqr(diff) + 0.5)
       sr2 = Int(Sqr(prod) + 0.5)
       If sr1 * sr1 = diff And sr2 * sr2 = prod Then
          Text1.Text = Text1.Text & a & Str(b) & Str(diff) & Str(prod) & crlf
       End If
     End If
 Next tot

 Text1.Text = Text1.Text & crlf & " done"
End Sub

  Posted by Charlie on 2015-12-03 15:28:25
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