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 Post subject: Math Essentials (Vol. O), Question #7.
PostPosted: Sat Sep 04, 2010 8:49 pm 
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Joined: Thu Jul 01, 2010 12:05 am
Posts: 35
The question asks: When integer 'i' is divided by 3, the remainder is 1. When 'i' is divided by 4, the remainder is 3. If 'i' is greater than 5 but less than 30, what is its value?

A) 13
B) 15
C) 16
D) 19
E) 23

The correct answer is "D)" and I got this one wrong by answering "A)". I was still unsure after consulting the answer sheet. The solution sheet gives a table with all the possibilities between two equations (3k + 1) and (4m + 3). The table shows different values until there appears to be a match. Additionally, there are separate columns for the values of "3k" and "4m". I don't understand how "3k" starts with a value of 6 and "4m" starts with a value of 4 when the integer 'i' is greater than 5 but less than 30. In short, I am confused as to why the table shows "3k" and "4m" columns and how we obtained 19 as the answer.

In my desperation once again, all I did for this question is to determine the Lowest Common Denominator (LCD) of 3 and 4, and I added 1 (i.e. lowest remainder) to the LCD thereby obtaining 13 as my answer.


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 Post subject: Re: Math Essentials (Vol. O), Question #7.
PostPosted: Mon Sep 06, 2010 6:44 am 
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Joined: Thu Feb 12, 2009 6:32 pm
Posts: 497
The different columns aren't really equations here. (THere's no equal sign, so they can't be.)
Instead we have the integer i divided by 3 with a remainder of 1. Thus, we start with an integer, multiply it by 3 and add 1 to it to arrive at i.
the integer (they're calling it k) multiplied by 3 is 3k. When we add1 to it, we get 3k+1. Those are the first two columns.
They they say that when i is divided by 4, we get a remainder of 3. Same process here. Take an integer, multiply it by 4, and then add 3 to it. Those are the third and fourth column.

THey start with value of k as 2. Multiply it by 3 and we get 6. Add one more and we get 7. That's a possible value for i. (We don't start with k=1 here because we'd be outside the acceptable range of "greater than 5 but less than 30")
Then start with value of m=1, multiply it by 4 and we get 4, and add 3 and we get 7. That's a possible value for i as well.
We need to find a situation where we have a value that works in both situations. 19 shows up both in the 3k+1 column and the 4m+3 column, so it's our answer.

Veritas Help


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