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It's not hard to figure out the problem algebraically...
Here Is A Math Trick
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?
It's not hard to figure out the problem algebraically...
thats very cool...and the progression/relationship is very obscure...some VERY smart person figured that one out...
OK showoff...
Actually, it only looks obscure. If you know the answer you want, the final equation is quite simple. Then back-track from there to come up with a funky-looking procedure to get there.
Show your work
that is a good way to get a girl's phone number.
that is a good way to get a girl's phone number.
Rack Jim
I think Jim's been racked quite a few times in his life.
Rack Jim
Assume your phone number is abc-defg where abcdefg are all digits from 0 to 9Show your work
abc = a*10^2 + b*10 + c
defg = d*10^3 + e*10^2 + f*10 + g
let X be the result from steps 2-5
X = ( abc * 80 +1 ) * 250 =
abc * 80 * 250 + 1 * 250 =
abc * 20000 + 250 =
abc * 2*10^4 + 250 =
2*(a*10^2 + b*10 + c) * 10^4 + 250 =
2 * (a * 10^2 * 10^4 + b* 10 * 10^4 + c * 10^4) + 250 =
2 * (a * 10^6 + b * 10^5 + c * 10^4) + 250
Let Y be the result from step 6
Y = X + defg = X + (d*10^3 + e*10^2 + f*10 + g)
= ( 2*(a * 10^6 + b * 10^5 + c * 10^4) + 250) + (d*10^3 + e*10^2 + f*10 + g)
Let Z be the result from step 7
Z = Y + defg =
(( 2*(a * 10^6 + b * 10^5 + c * 10^4) + 250) + (d*10^3 + e*10^2 + f*10 + g) ) + defg =
( 2*(a * 10^6 + b * 10^5 + c * 10^4) + 250) + (d*10^3 + e*10^2 + f*10 + g) + (d*10^3 + e*10^2 + f*10 + g) =
( 2*(a * 10^6 + b * 10^5 + c * 10^4) + 250) + 2*(d*10^3 + e*10^2 + f*10 + g) =
2*(a * 10^6 + b * 10^5 + c * 10^4) + 250 + 2*(d*10^3 + e*10^2 + f*10 + g) =
2*(a * 10^6 + b * 10^5 + c * 10^4) + 2*(d*10^3 + e*10^2 + f*10 + g) + 250 =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) + 250 =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) + 250
Let A be the result from step 8
A = Z - 250 =
(2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) + 250) - 250 =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) + (250 - 250) =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) + (0) =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g)
Let B be the result from step 9
B = A/2 =
2 * (a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g) / 2 =
a * 10^6 + b * 10^5 + c * 10^4 + d*10^3 + e*10^2 + f*10 + g =
abcdefg
i.e., your original phone number
Thats what I was gonna say
That's like that name sex number formula...
I sent this to a friend a few weeks back and she didn't get it. I guess she doesn't even know how to use a calculator. She used one but didn't get it. She's not as smart as Travis.
You need to get out of here!!!!
You're too smart for this forum
damnit. thank you, now my head hurts againi just finshed doing my math.
Where's Blaze? :p
Absolutely correct. But in this case there's an easier way.
Instead of splitting the phone number up into 7 distinct digits, split it into two numbers...the three-digit exchange (call this 'a') and the four-digit extension (call this 'b'). As we shall see, the only necessary condition for this "trick" to work is that b<10000.
Expressing the original problem as an algebraic expression, we get:
x = (250*(80*a + 1) + b + b - 250) / 2
Simplifying gives us:
x = (20000*a + 250 + 2*b -250) / 2
x = (20000*a + 2*b) / 2
or
x = 10000*a + b
In other words, the first part of the phone number is shifted 4 places (with trailing zeros added) and then the second part is added.
As such, this would even work if you augmented the original phone number with the area code (for example, (210)555-1212 would split into a=210555 and b=1212).
Do you program shaders? hehe
... yeah... but we already know the steps...
math. Us english nerds just care about learning enough to get correct change back when we buy beer at HEB.
You are an animal.
Whoever has time to think of that equation should be shot on site.
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