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Simple proof |
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| Feb11-04, 04:33 PM | #1 |
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Simple proof
This should be an easy question but I'm having problems with it. Prove that any number that ends in five when squared equals 25. So if n is the number then
(n/5)^2 = (n^2)/25 Although if you expand the left side then this statement is redudant. Can someone help me with this? |
| Feb11-04, 04:51 PM | #2 |
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I think you ought to reread the question - 15*15 ends in a five, do you mean if x is divisibly by 5, then x^2 is divisible by 25?
well, 5|x implies x=5y some y, so x^2=25y^2, so 25 divides x^2 is a formal statement of it. |
| Feb11-04, 04:54 PM | #3 |
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any positive number that ends in 5 when squared ends in 25
eg 5^2 = 25 15^2 = 225 25^2 = 625 Just scrap what I started with I don't think it helps at all, how could I prove this question? |
| Feb11-04, 04:56 PM | #4 |
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Simple proof
oh, ok
ends in 5 is the same as is equal to 10r+5 for some r safely we can leave the rest to you |
| Feb11-04, 05:15 PM | #5 |
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I wouldn't say safely could you please expand on that? every time a number that ends with 5 is squared the resulting term ends in 25
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| Feb11-04, 05:20 PM | #6 |
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square 10r+5 you get a 25 and something that is a multiple of 100.
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| Feb12-04, 05:53 AM | #7 |
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matt grime was making the perhaps unwarrented assumption that a person asking such a question could do basic algebra.
(10r+ 5)2= 100r2+ 2(10r)(5)+ 25 = 100r2+ 100r+ 25 = 100(r2+r)+ 25 Because r2+r is multiplied by 100, 100(r2+r) will have last two digits 00. Adding 25 to that, the last two digits must be 25. |
| Feb12-04, 06:14 AM | #8 |
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I was hoping that given the start the questioner would work on the answer some more and get the solution themselves. Don't know about you, Halls (if I can be familiar ;-)) but a lot of the queries appear to me to be from homework sheets; is it better to prompt the right answer or spoonfeed it verbatim?
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| Feb12-04, 10:52 AM | #9 |
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Ya this is the property which is applied in vedic maths
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| Feb13-04, 07:24 AM | #10 |
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Actually, Matt, I was being sarcastic. You had given very good answers and the orginally poster repeatedly asked for more.
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