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 Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 January 22nd, 2013, 02:42 AM #1 Newbie   Joined: Jan 2013 Posts: 2 Thanks: 0 Math Possibilities Hello everyone. MY first time here Hope someone can help me this problem. Peter has fewer than 1000 cars. With that in the large parking lot he made 5 equal rows of cars and have one car left. He also made 4 equal rows of cars and also have one car left. He then made 9 equal rows of cars and didnt have any cars left. He can have six possibilities. So how many cars does he have? Thank you all , Marta
 January 22nd, 2013, 07:11 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 12,756 Thanks: 860 Re: Math Possibilities 81 or 261 or 441 or 621 or 801 or 981
 January 22nd, 2013, 09:34 AM #3 Newbie   Joined: Jan 2013 Posts: 2 Thanks: 0 Re: Math Possibilities Denis thank you so much. But would you pls explain how you arrive to those numbers. Was there a formula used, short cut or how?
January 22nd, 2013, 02:49 PM   #4
Math Team

Joined: Dec 2006
From: Lexington, MA

Posts: 3,267
Thanks: 407

Re: Math Possibilities

Hello, Marta!

Welcome aboard!

Quote:
 Peter has fewer than 1000 cars. In the large parking lot he made 5 equal rows of cars and had one car left. He also made 4 equal rows of cars and also had one car left. He then made 9 equal rows of cars and didn't have any cars left. How many cars does he have?

$\text{Let }N\text{= number of cars.}$

$N\text{ divided by 4 leaves a remainder of 1.}$
$N\text{ divided by 5 leaves a remainder of 1.}$
$\text{Hence: }\:N \:=\:20a\,+\,1\,\text{ for some integer }a.$
[color=beige]. . [/color](Do you see why?)

$N\text{ divided by 9 leaves no remainder.}$
$\text{Hence: }\:N \:=\:9b\,\text{ for some integer }b.\;\;[1]$

$\text{W\!e have: }\:9b \:=\:20a\,+\,1 \;\;\;\Rightarrow\;\;\;b \:=\:\frac{20a\,+\,1}{9} \;\;\;\Rightarrow\;\;\;b \:=\:2a\,+\,\frac{2a\,+\,1}{9}\;\;[2]$

$\text{Since }b\text{ is an integer, }\,2a+1\text{ must be a multiple of 9.}$
[color=beige]. . [/color]$\text{This happens when: }\,a \:=\:4,\,13,\,22,\,31,\ldots$
$\text{That is, when }a \:=\:9m\,-\,5$

$\text{Substitute into [2]: }\:b \:=\:2(9m\,-\,5)\,+\,\frac{2(9m\,-\,5)\,+\,1}{9} \;=\; 20m\,-\,11$

$\text{Substitute into [1]: }\: N \:=\:9(20m\,-\,11) \;\;\;\Rightarrow\;\;\;N \:=\:180m \,-\,99$

$\text{Therefore:}$

[color=beige]. . . [/color]$\begin{array}{c|c} m & N \\ \hline \\ 1 & 81 \\ \\ \\ 2 & 261 \\ \\ \\ 3 & 441 \\ \\ \\ 4 & 621 \\ \\ \\ 5 & 801 \\ \\ 6 & 981 \\ \\ \hline \end{array}$

 January 23rd, 2013, 08:02 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,821 Thanks: 1047 Math Focus: Elementary mathematics and beyond Re: Math Possibilities N = 5a + 1 N = 4b + 1 0 = 5a - 4b, a/b = 4/5 so b is a multiple of 5. Starting at 20, add 45 (smallest multiple of 5 divisible by 9): b, 4b + 1 20, 81 65, 261 110, 441 155, 621 200, 801 245, 981.

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