I am having trouble with this one.... Prove that any positive integer whose ALL digits are 1s (except 1) is not a perfect square.
I didn't mean 11^8 either, I meant 11 written in base 8. For example, 34 written in base 8 is equal to 3 * 8^1 + 4 * 8^0 = 28 in base 10.
I think it should be: [tex] N = \sum_{n=1}^\infty (1*10^{n-1}) [/tex] btw, it should be base 10 then.
i think it should be: [tex] N = \sum_{n=1}^k (1*10^{n-1}) [/tex] where k can be any positive integer.
suppose x^2 = 1+10+100+... then mod 10, x=+/-1, that is x = 10r+1 or 10r-1 for some r. consider (10r+1)^2 = 100r^2+20r+1 this is equal to 1+10+100+... assume r is not zero, 100r^2+20r=10+100+... divide by 10, 10r^2+2r=1+10+.. but the rhs is even and the lhs is odd, so it must be that r=0. similarly we reason for x=10r-1 and decide r=0 too.
how come the RHS is even and LHS is odd?? i don't quite understand this stuff, can you recommend me a book or an article that is simple to learn?
my apologies, i meant to say the lhs is even (it is divisible by two) and the rhs is odd since it is 1 plus a multiple of 10
Explanation of matt's proof : x^2 = 1+10+100+...=1+ 10(1+10+...)=1+10k, say, where k=1+10+..., the number within (). x^2 = 1+10k means that the units digit of x^2 is 1 (example, 81, 121, etc.) Now for the units' digit of x^2 to be 1, the units digit of x must be 1 or 9. (example : 11^2 =121, 19^361, but 23^2 must end with 9 and 45^2 ends with 5 etc.) If the units' digit of x is 1, this means that x can be written as 1 + 10a + 100b +... = 1 + 10r (example : 151 = 1 + 50 + 100 = 1 + 10*15 ) If the units' digit of x is 9, this means that x can be written as 9 + 10a + 100b + ... (10-1) + 10a + 100b +... = 10 + 10a + 100b + ... -1 = 10r -1 (example 69 = 70 - 1 = 7*10 - 1) Having shown that x must be of the form 10r + 1 or 10r - 1, the rest is just algebra...and is easy to follow in matt's proof. In the last line, you have 10r^2 + 2r = 1 + 10 + ... LHS = 2*5r^2 + 2r = 2*(5r^2 + r) = 2n, an even number RHS = 1+10+100+... = 1 + 2(5 + 50 + ...) = 2n + 1, an odd number
So, if the LHS is an even number and RHS is odd, it should not be a perfect square?? if not please explain the very basics.
i seem to have another proof, every odd perfect square when divided by 8 has a remainder of 1... and noting that if an odd integer is squared the answer is always odd. and with the exception of 1,11, every positive integer whose all digits are 1 have a remainder of 7. But 11 is clearly not a perfect square. I think this should be enough proof.. but please explain Matt Grime's proof, because I'd like to learn.
matt's proof is known as reductio ad absurdum or proof by contradiction. You make an assumption (in this case, you assume that some perfect square can be a number of the form 11...1) and then arrive at a logical contradiction (such as the statement "even number = odd number"). This implies that the assumption that was made was incorrect (so it should not be possible to find any perfect square of the form 11...1), and that's what you wanted to prove.
This is a perfectly good proof, as long as you show how these statements are true. They are true, no doubt...but I would imagine that you would have to show, for instance that odd perfect squares leave residues of 1 when divided by 8. You might also have to prove that 11...1 always leaves a remainder of 7 when divided by 8. The first is proved as follows : N=(2k+1), say, some odd integer. N^2 = 4k^2 + 4k + 1 = 4k(k+1) + 1 Since k and k+1 are consecutive numbers, one of them must be even. So their product must also be an even number. So k(k+1) = 2m, say. Then N^2 = 4*2m + 1 = 8m + 1. This last equation tells you that N^2, when divided by 8, gives a remainder of 1. To prove that 11...1 leaves a remainder of 7, do the following : i. Write this as the sum of powers of 10, as in post#8 ii. What happens when you divide 1000 by 8 ? And higher powers of 10 ? iii. This argument leaves you with having to find the remainder for only 111, which is 7 (all other numbers of the form 11...1 have the same remainder as 111). Now you are really done. For a 15 year old, your math skills are far removed from "pretty crap."
The answer to this question Tn=a+(n-1)d a=11 ......n d=10 ....................1 T1=11+(1-1)10 = 11 ....................2 T2=11+(2-1)10 = 111 ....................3 T3=11+(3-1)10 = 111 . . . . . Tn=a+(n-1)d ....................................2 To be a perfect square = x = Tn 11 is not a squared number and it is an odd number so (a=11) is not a squared number and it is an odd number. ......................................................................n So (n-1)d is always an even number because ( d=10 ) is always an even number So (n-1)d is always an even number So if you add an even number with an odd number the answer is always an odd number. So Tn is always an odd number. So an odd number can never be a square. Even though this could be obvious once you read the question this is the way to prove that all the positive integers where digits are 1s (except 1) is not a perfect square.
Supundika, This is NOT the way to prove this. You proof is fatally flawed. An odd number can never be a square ?? How about 1 or 121 ? Also, the very basis of your argument is incorrect. First of all, d is really d^n, but even that doesn't fix it. T(3) = 11 + 2*1000 = 2011, definitely not 1111 . So you start off wrong, and even if you didn't, all you would be proving is that 111..1 is an odd number : a rather convoluted way to prove something that simple. I suggest you be more sure of your math before you go about proclaiming the superiority of your faulty proof over others previously provided.
Gokul43201, Your skill in mathematics is really impressive;it is a fact whether i envy it or not. it is good to point out other's mistakes but i am sorry i have to say that your criticism about Supundika is not nice at all.