1111 - 22
Here's a fairly straightforward one:
Consider the sequence of numbers: 11-2, 1111-22, 111111-222,...
Show that they are perfect squares. What else do you notice?
Consider the sequence of numbers: 11-2, 1111-22, 111111-222,...
Show that they are perfect squares. What else do you notice?
Labels: mathemagic
posted by Chris at
5:39 PM
11 Comments:
11 -2 = 9 = 3 x 3
1111 -22 = 1089 = 33 x 33
111111 -222 = 110889 = 333 x 333
11111111 -2222 = 11108889 = 3333 x 3333
1111111111 -22222 = 1111088889 = 33333 x 33333
11 -2 = 9 = 3 x 3
1111 -22 = 1089 = 33 x 33
111111 -222 = 110889 = 333 x 333
11111111 -2222 = 11108889 = 3333 x 3333
1111111111 -22222 = 1111088889 = 33333 x 33333
111111111111 -222222 = 111110888889 = 333333 x 333333
I suppose the sequence keeps going
11-2 = 1*(11-2) = 3^2
1111-22 = 11*(101-2) = 11*99 = 11^2 * 3^2
111111-222 = 111*(1001-2) = 111*999 = 111^2 * 3^2
11111111-2222 = 1111*(10001-2) = 1111*9999 = 1111^2 * 3^2
et cetera
Call the sequence of differences A-B
The first term A is always a repdigit of 2n ones, i.e.
n=1 A = 11 = 10^1*1 + 1*1
n=2 A = 1111 = 10^2*11 + 1*11
n=3 A = 111111 = 10^3*111 + 1*111
Call the repdigit of "n" ones R_1
and the repdigit of "n" nines R_9
then for any n, A = (10^n+1)*(R_1)
A = (R_9+2) * (R_1)
A = (9 * R_1 + 2) * R_1
The second factor B is always a repdigit of "n" twos, call it
R_2 = 2*R_1
So the difference is
A-B
= (9*R_1 + 2) * R_1 - 2 * R_1
= (9*R_1 + 2 - 2) * R_1
= 9* R_1 * R_1
= (3*R_1)^2
Thus the difference always equals a square.
If there is "something else" I should notice, it's probably what I said above about A and B....
Ragknot. If you're using IE8, then turn compatibility mode on, or try restaring your PC. I didn't say it was a ToM.
You both got it. Thanks for going the extra mile Ross.
Here's my pre-prepared solution:
111111 - 222 = 11100 + 111 - 222 = 111000 - 111 = 111(1000-1)
= 111*999 = 111*111*9 = (111*3)² == 333²
That was for n = 3. Now do again, but imagining n arbitrary.
What's "ToM"?
ToM = Trick of Mind.
Problems often evolve a "trick" solution.
(10^n-1) is a repdigit of n nines
(10^n-1)/9 is a repdigit of n ones
thank you, Wolfram
It works for any single digit d:
(d/9)*(10^n-1) is a repdigit of n d's
The only trick I can see was recognizing the hidden connection between the two repdigits.
A perfect square is the square of an integer; I guess an imperfect square is a number that is the square of a non-integer, like 2 is the square of 1.414...
For clarification, although mathematicians use the term "square [number]", they implicitly mean an integer squared. They used to say "perfect square" to emphasise this. If a number is not a perfect square, then it is an imperfect square.
The "thing to notice" was that all the square were the squares of a string of 3s.
Ross. Thank for that repdigit stuff. I think that's pretty good.
So good, that I'm going to solve the problem with it.
We have dddd...ddd = (d/9)(10^n - 1), where d is repeated n times.
The original problem then is:
(1/9)(10^(2n) - 1) - (2/9)(10^n - 1)
= (1/9)(10^n - 1)(10^n + 1) -(2/9)(10^n - 1)
= (10^n - 1)(10^n + 1 -2)/9
= ((10^n - 1)/3)^2
And so is a perfect square.
= ((3/9)(10^n - 1))^2
= (333...333)^2, where 3 is repeated n times.
A bit late, but anther useful trick is x.abcdabcdabcdabcd...
=x + (abcd/9999) etc.
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