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A four digit number

With YummyCookies' permission I am posting this problem:

If N = aabb is a perfect square, find N. Detailed calculations must be shown. Good Luck. R.B.

759 day(s) ago

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Answers (3)

jcatta21

Based on my knowledge of perfect squares, the square would have to be some where between the nearest thousands. The list of perfect squares in the numeric system are as follows:
4 = 2^2
16 = 4^2
25 = 5^2
36 = 6^2
49 = 7^2
64 = 8^2
81 = 9^2
100 = 10^2
121 = 11^2
144 = 12^2
169 = 13^2
196 = 14^2
225 = 15^2
256 = 16^2
289 = 17^2
324 = 18^2
361 = 19^2
400 = 20^2
441 = 21^2
..................................
(x + 10n)^2 = x^2 + nâˆ‘(n â€“ 1) 20x + 100 (2n â€“ 1).

Using my formula for squaring any number to find N = aabb, this can be proven via the first square yielding four digits, which would be 32^2 = 2^10 = 1024. This perfect square can serve as a starting point for computing the square of a number with duplicate double digits. Using my formula for squaring any number yields at most number starting from 80 in the inclusive range between 32 && 99:

(x + 10n)^2 = (x + 10(n - 1))^2 + 20x + 100(2n - 1)
(1 + 10(8))^2 = 1^2 + [[20(1) + 100(2(8) - 1] + 20(1) + 100(2(7) - 1) + 20(1) + 100(2(6) - 1) + 20(1) + 100(2(5) - 1) + 20(1) + 100(2(4) - 1) + ... + nâˆ‘(n â€“ 1) 20x + 100 (2n â€“ 1)] = 6481
Using the above formula as a basis and the fact that n^2 = (n - 1)^2 + 2n - 1 yields the succeeding squares:
(2 + 10(8))^2 = n^2 = (n - 1)^2 + 2n - 1 = 80^2 + 161
.....................................
(8 + 10*8)^2 = 8^2 + [

20(8) + 100(2(8) - 1)
20(8) + 100(2(7) - 1)
20(8) + 100(2(6) - 1)
20(8) + 100(2(5) - 1)
20(8) + 100(2(4) - 1)
20(8) + 100(2(3) - 1)
20(8) + 100(2(2) - 1)
20(8) + 100(2(1) - 1)
] = 7744

Thus, all perfect squares that are in the inclusive range between 1024 && 10000 have at most one perfect square 7744 = 88^2.

J.C

(x + 10n)^2 = (x + 10(n - 1))^2 + 100 (2n - 1)

Posted 756 dy ago

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math_fanatic

The square root of 7744 is 88.

I don't know what detailed calculations you want us to show.
I just used a calculator and started at 1100 and took the square
root of all the numbers that fit the description until I got one that was
a perfect square. There are only 81 of these numbers so I didn't
think it was necessary to come up with a clever algorithm to find the one
that was a perfect square. Is there another way to do it? Was there
some formula you wanted us to use?

Posted 759 dy ago

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Vince

Let X = the number which when squared will give the perfect square N = aabb

Therefore X^2 = 1000*a + 100*a + 10*b + b = 1100*a + 11*b = 11(100*a + b)

therefore X^2/11 = 100*a + b. So, X must be divisible by 11

that leaves 11, 22, 33, 44, 55, 66, 77, 88, and 99 as the only possible choices. But 11, 22, or 33 cannot be it because 33^2 /11 < 100 and 100*a + b must be at least 100. Same goes for 11 and 22.

therefore 44, 55, 66, 77, 88, or 99 are the only remaining choices.

easy solving for X^2/11 with the remaining choices leaves 88 as the only number that satisfies X^2/11 = 88^2/11 = 704 = 100*a + b = 100*7 + 4, therefore the answer is N = 7744.

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