There exists a number oddity with 3 different 4-digit numbers. One is 9801, where (98 + 01)^2 = 9801. It also works with 3025: (30+25)^2 = 3025.
What is the other number?
What is the smallest 6-digit number that would work?
(in other words, in a 6-digit number abcdef: abcdef=(abc+def)^2)

Expressing the problem algebraically,<o:p></o:p>

<o:p> </o:p>

(a+b)^2 = 100a + b            (1)<o:p></o:p>

<o:p> </o:p>

This can be written as:<o:p></o:p>

<o:p> </o:p>

a^2 + (2b-100)a + (b^2-b) = 0<o:p></o:p>

<o:p> </o:p>

Solving for a:<o:p></o:p>

<o:p> </o:p>

a = 50-b ¡À ¡Ì(2500-99b)<o:p></o:p>

<o:p> </o:p>

We need values of b that yield an integer after taking the square root.  There are probably elegant ways to do this, but I just used an Excel spreadsheet.  We know from the square root term that b can¡¯t exceed 25, so it¡¯s pretty easy to find that 1 and 25 are the only values of b that work.<o:p></o:p>

<o:p> </o:p>

For b=1:<o:p></o:p>

<o:p> </o:p>

a = 49 ¡À 49 = 98 choosing the + sign<o:p></o:p>

<o:p> </o:p>

Eq (1) thus gives (98+01)^2 = 100 x 98 + 1 = 9801<o:p></o:p>

<o:p> </o:p>

For b=25:<o:p></o:p>

<o:p> </o:p>

a = 25 ¡À 5 <o:p></o:p>

<o:p> </o:p>

Choosing the + sign gives a=30 and 3025<o:p></o:p>

<o:p> </o:p>

Choosing the ¨C sign gives a=25 and 2025 for the answer to the problem.<o:p></o:p>

<o:p> </o:p>

A similar approach yields the six-digit number 494209.

Posted by NK on 2004-12-09 16:32:07