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2006-06-29 00:17:16 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Cannot be, because any whole number ending on a 1 squared will end on a 1.

Did you mean the square root of 12345678987654321?

This results from summation:

111111111^2 multiplied manually is

11111111100000000
01111111110000000
00111111111000000
00011111111100000
00001111111110000
00000111111111000
00000011111111100
00000001111111110
00000000111111111

12345678987654321

2006-06-29 00:42:29 · answer #1 · answered by jorganos 6 · 1 0

It is to the 3rd decimal place, but but not after that:

Sqrt(123456789) = 11111.111060555556440..

But that's interesting.

Note: 11111^2 = 123454321

That's another interesting one, but this time we can see why. We can express 11111 as

(10000+1000+100+10+1)^2

We can see the last digit will be 1
We'll only have 2 tens: 1*10+1*10= 2*10
For hundreds we'll have 2*100+10*10 = 3*100
thousands: 2*1000+2*10*100 = 4*1000
10 thousands: 2*10000+2*10*1000+100*100= 5*10000
etc.
Because all digits are 1 and the power is only 2, we get a simple symmetric digit distribution.
As the others have pointed out this true for strings of 1s up to 10
(1111111111)^2 = 12345678900987654321
which is symmetric but breaks down at 11 1s
(11111111111)^2 = 123456790120987654321
This is not symmetric since number of 1s is greater than the number of digits (0 to 9) that can appear at a given place in the number, and so there is a carry over.

There is a range of numbers that when we take the square root of them, all digits to the left of the decimal place will be 1:

sqrt(123454321) =11111.000...
sqrt(123456790)=11111.11110555555 (1 more 1)
sqrt(123476543) = 11111.99996
sqrt(123476544) = 11112

2006-06-29 08:30:42 · answer #2 · answered by Jimbo 5 · 0 0

Yes dear!
Try to get the squares of the "all 1 numbers" (1, 11, 111, 1111, 11111, etc.)
You will find something amassing!
1 1
11 121
111 12321
1111 1234321
11111 123454321
111111 12345654321
1111111 1234567654321
11111111 123456787654321
111111111 12345678987654321
See the sequence of the resulting numbers. Also, the middle number is how many aces we have. Thus it goes from one to the number of ones we take the square and drops to one again!

2006-06-29 07:45:30 · answer #3 · answered by soubassakis 6 · 0 0

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