take any 3digit number x and four digits y
the "phone number" is x*10000 + y
your formula is
((x*80 + 1)*250 +y +y -250)/2
= x*80*250/2 +2y/2
= x*10000 + y
so it has to work
2007-04-13 09:14:04
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answer #1
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answered by hustolemyname 6
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Let's assign the digits of the phone number to variables.
1st three digits = x and last four digits = y
Now:
.5(250*(80x + 1) + 2y - 250)
.5(20000x + 250 + 2y - 250)
.5(20000x + 2y)
10000x +y
As phone numbers are seven digits, the three digit number creates a number with with those three digits in the millionth, hundred thousandth, and ten thousandth places (in order, as 1*x=x). Therefore, the four digits represented by y go into the thousandth, hundredth, tenth, and ones places (in order, as 0+y=y and there are no digits in those places).
2007-04-13 16:19:20
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answer #2
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answered by infinitys_7th 2
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((80x + 1)*250 + 2y -250)/2
=(20000x+250+2y-250)/2
=10000x+125+y-125
=10000x + y
Multiplying x by 10000 yields the three digit x value along with 4 trailing zeros. Adding y replaces those zeros with the last four digits of your phone number.
2007-04-13 16:20:47
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answer #3
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answered by Nicholas V 2
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Let the 1st 3 digits be A, and remember they're in the 10,000's place; let the last 4 be B. So your number is 10,000A + B.
The directions say:
[(80A + 1)250 + 2B - 250] / 2 =
[20,000A + 250 + 2B - 250] / 2 =
[20,000A + 2B] / 2 =
10,000A + B
2007-04-13 16:13:45
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answer #4
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answered by Philo 7
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Let x be the first three digits of my phone number, and y be the rest.
Let d be the value displayed.
2. d=x
3. d=80x
4. d=80x+1
5. d=250(80x + 1) = 20000x + 250
6. d=20000x + 250 + y
7. d=20000x + 250 + 2y
8. d=20000x + 2y
9. d=10000x + y
2007-04-13 16:16:34
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answer #5
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answered by Amit Y 5
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step 1 -pick a number -
step 2 -double it -
step 3 -add 10 -
step 4 -divide it by 2 -
step 5 -subtract the number you picked in step 1
You get 5
2007-04-13 16:17:15
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answer #6
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answered by Anonymous
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wow thats cool how that works how did you know about this problem
2007-04-13 16:14:45
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answer #7
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answered by jazze pha 1
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