Fishing with Rod Discussion Forum
Fishing in British Columbia => General Discussion => Topic started by: The Gilly on January 24, 2005, 11:12:48 AM
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1. Grab a calculator. (you won't be able to do this one in your head)
2. Key in the first three digits of your phone number (NOT the area >code)
3. Multiply by 80
4. Add 1
5. Multiply by 250
6. Add the last 4 digits of your phone number
7. Add the last 4 digits of your phone number again.
8. Subtract 250
9. Divide number by 2
Do you recognize the answer?
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That is neat. Also, if you abstract it, you can see why it works:
If you use a varialbe, x, for the first 3 digits and a variable, y, for the last four digits, you can write an expression and simplify it:
80x
80x+1
250(80x+1)
250(80x+1)+2y
250(80x+1)+2y-250
[250(80x+1)+2y-250]/2
Factor 2 out of the numerator:
2[125(80x+1)+y-125]/2
Cancel out 2:
125(80x+1)+y-125
Distribute:
10000x+125+y-125
Simplify:
10000x+y
When you multiply the first 3 digits by 10000 you will have xxx0000 and then you add the last four digits to end up with xxxyyyy, your telephone number! Therefore it should work no matter what the phone number.
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;D ;D ;D and thanks ds for the high school maths class flash backs arghhhh lol ;)
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LOL - math is fun! :o
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LOL - math is fun! :o
I hated math and still do except when you are measuring the lenght and the weight of a fish. Go Imperial not Metric, Imperial sounds a lot bigger. ;D ;D
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you guys need to go fishing
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you guys need to go fishing
Taking your advice and going tomorrow and will get that first hatchery as well. So much for the math. ;D
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New math question. If you co fishing 0 times for Steelhead, how many will you catch? :'( :'( :'( :'( :'(
Somebody call my wife and plead my case (http://smileys.smileycentral.com/cat/4/4_2_111v.gif)
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Well, being the wife of a fisherman (and mother of 4 kids), I have to admit I do sympathize with your wife ;)
but I always tell myself, there are much worse things a husband could be doing besides fishing. Maybe just run that by her
Don't know if that will help your case at all ;D
Cool math question btw, (just so it doesnt look like I am high-jacking the thread :))