Question 1208966: the numbers on the houses on the north side of carlito street are consecutive odd starting with 1. The plastic digits used to label each house are $0.04 per digit. If it costs $58.12 to label all the houses on the north side of Carlito street, how many houses are there?
Found 3 solutions by ikleyn, mccravyedwin, Alan3354: Answer by ikleyn(52814) (Show Source):
You can put this solution on YOUR website! .
The numbers on the houses on the north side of Carlito street are consecutive odd starting with 1.
The plastic digits used to label each house are $0.04 per digit. If it costs $58.12 to label
all the houses on the north side of Carlito street, how many houses are there?
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The number of the digits bought is = 1453.
Of them, 5*1 = 5 digits are for five 1-digit odd numbers from 1 to 9; 1453-5 = 1448 digits left.
45*2 = 90 digits are for 45 2-digit odd numbers from 11 to 99; 1448-90 = 1358 digits left.
450*3 = 1350 digits are for 450 3-digit odd numbers from 101 to 999; 1358-1350 = 8 digits left.
These 8 digits are for two the 4-digit odd numbers 1001 and 1003 on the north side of Carlito street.
The total number of houses on the north side of Carlito street is 5 + 45 + 450 + 2 = 502. ANSWER
Solved.
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Below is my comment regarding Edwin' solution to this problem.
The solution by Edwin, giving the answer 501 for the number of houses
with odd plates on them - is INCORRECT.
The Edwin's error is in the last two sentences in his post before the word Answer :
he incorrectly counted even numbers from 0 to 1002, inclusively,
and incorrectly counted odd numbers from 1 to 1003, inclusively.
The correct count for even numbers/plates from 0 to 1002, inclusively, is 502,
as well as the correct count for odd numbers/plates from 1 to 1003, inclusively, is 502.
Answer by mccravyedwin(408) (Show Source):
You can put this solution on YOUR website!
Let's pretend they put plastic digits on ALL the houses on both sides of the
street. and that there are the same number of houses on both sides of the
street.
The number of digits in all the even integers 0,2,...,2n is the same as the
number of digits in all the odd integers 1,3,...,2n+1
It would have cost twice as much, $58.12x2 = $116.24, to put plastic digits on
all the houses on both the north and south sides of the street, that is, IF the
first even-numbered house were numbered 0.
[Even though 0 is not normally used as a house number, let's pretend it is
on this street anyway.]
If we consider 0 as a 1-digit number, then
there are 10 1-digit numbers which would cost $0.04x10=$0.40,
there are 100-10=90 2-digit numbers which would cost $0.04x2x90=$7.20,
and there are 1000-100=900 3-digit numbers which would cost $0.04x3x900=$108.
That would cost $0.40+$7.20+$108=$115.60
Since $116.24-115.60=$0.64, that means they placed $0.64/$0.04 = 16 more
plastic digits than they placed on houses numbered with 3 or fewer digits.
So 16 plastic digits were placed on 4 houses with 4-digit numbers.
These are the houses numbered 1000, 1001, 1002, and 1003.
So the even house numbers are 0,2,4,...,1002 and there are 1002/2=501 of them.
The odd house numbers are 1,3,5,...,1003 and there are also 501 of them.
Answer: There are 501 odd numbered houses on the north side of the street.
Edwin
Answer by Alan3354(69443) (Show Source):
You can put this solution on YOUR website! Not a solution the this problem, but:
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Our address at work is 910 Industrial Blvd.
A guy game in, asked for help finding his customer.
Asked me where 909 Industrial was.
Obviously, it was directly across the street.
I told him that, went out front and pointed to it.
He was grateful.
He was driving a "big rig," 18 wheeler.
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How can a person making deliveries for living not know about odd and even addresses?
I considered explaining that to him, but decided not to.
Probably then, "How do you know if an address is odd or even?"
There is no lower limit to intelligence.
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