... digits ... there ... and ... then ... OK, If you have n+1 numbers, 1,11,111,1111,11111,111111,...., two of them must have the same residue mod n, the...
... another ... of ... I ... of ... of ... Sorry I forgot to put the "hole" in the "pigeon". By the way,I think this "principle" is quite powerful. For...
... lemma: a power of 3 greater than 0 and ending in XYZ, each X,Y,Z being a digit from 0 to 9 must uniquely follow a power of 3 that has the 3 digit ending...
... <julien_santini@> ... does not Here is, I think, another proof: The only numbers whose squares end in a 1 are those with last digits 9 or 1. So, the...
Beats me... perhaps some examples would make it more clear. You said " ....... lemma: a power of 3 greater than 0 and ending in XYZ, each X,Y,Z being a digit...
... Then I mentioned that -m could be substituted for the value of -1 within each of the parenthesis. ... Still other variable "b" and "w" to add to the mix. ...
... I agree that some more examples might help some to better understand. Not all endings can be a power of three, "000,002,004,005,006,008,010,..." which are...
... The sum of the geometric series 10^p + 10^(p-1) + 10^(p-2) + . . . + 1 constitures a number of all 1's. The sum of this series can also be written as...
... You can modify your proposition and make it more general. "A number whose LAST TWO DIGITS ARE ONES can not be a perfect square" Demonstration: Consider a...
... Yikes! ... Should have been The sum of the geometric series 10^(p-1) + 10^(p-2) + . . . + 1 I quiclly remember the sum of a geometric series by adding a...
... minus sign on b, and for ... 10^10), a=0. For ... as the ... I just realized that I can prefix a letter to the number to indicate the number of exponents...
Prove in more than one way that a isoceles right triangle cannot have all integer sides. I can think of 2 distinct proofs. there may be more. Enjoy, Cino...
Since, in this case, the hypothenuse will always be a side times square root of 2, which is an irrational number. It would be impossible for all sides to be...
... all ... 1. Since the hypotenuse is always longer than either leg, that means the lengths of the 3 sides are x, x, and sqroot(2x^2) = x*sqroot(2) So...
... have all ... (Sorry my last answer was a duplicate of somebody else's... for some reason his answer had not yet made it to my email.) Here's another way: ...
So far, we have assumed sqrt(2) is irrational. Perhaps we need to show this is the case *after* we prove the assertion which can also be stated more generally...
... indicate ... letters, ... would ... I am not sure what you mean, Please give examples for your notation to express the following: 123456 23456789 ...
In a message dated 8/8/2007 3:59:19 PM Central Daylight Time, ... I don't think so, only that it's not an integer. stevo </HTML> [Non-text portions of this...
In a message dated 8/8/2007 8:04:04 PM Central Daylight Time, ... A5'123456 A7'23456789 B1'14'1234567 12345678 (the space is not part of the notation) Letting...
You are saying that that sqrt(2) is NOT irrational? If this is so, I gotta go back to study basic math, as I've always read it in all math books that sqrt(2)...
In a message dated 8/9/2007 4:43:57 PM Central Daylight Time, ... No, of course not. I'm saying that the assumption is only that sqrt(2) is not an integer....
... What are the possibilities of the sqrt(2)? 1. It is a rational number 2. it is an irational number If we assume 2., then we also assume it is not an...
... Not so. If the square root of two is rational then for integer sides we make the two equal sides of the isoceles triangle equal to a multiple of the...
In one old Math Competition I found the following problem, but I could not find a solution. Could some member of the Group help me to solve it ? The problem...
... could ... a ... maybe do powers of 3 mod 1,000 and look for a number which is next to a number divisible by 2^3 and a number divisible by 5^2. e.g. 023,...
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