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,
(a+b)^2 = 100a + b (1)
This can be written as:
a^2 + (2b-100)a + (b^2-b) = 0
Solving for a:
a = 50-b } (2500-99b)
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.
For b=1:
a = 49 } 49 = 98 choosing the + sign.
Eq (1) thus gives (98+01)^2 = 100 x 98 + 1 = 9801
For b=25:
a = 25 } 5
Choosing the + sign gives a=30 and 3025
Choosing the - sign gives a=20 and 2025 for the answer to the problem.
A similar approach yields the six-digit number 494209.
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Posted by NK
on 2004-12-09 21:26:46 |
And Again
