Why can x only be a number in "35n + 17"

Question

r => row of chairs

x => total amount of chairs

 

When you try to make 5 rows, there's 12 chairs left over.

When you try to make 7 rows, there's 4 chairs too little.

 

And that means:

"x - 12" must be evenly divisible by 5; 

"x + 4" must be evenly divisible by 7;

 

Find what x´s are possible

---------------

Numbers that are after subtracted by 12 evenly divisible by 5: 

17, 22, 27, 32, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87...

 

Numbers that are evenly divisible by 7 after 4 is added to that number:

3, 10, 17, 24, 31, 38, 45, 52, 59, 66, 73, 80, 87...

 

So as you can see, the numbers that fit into both the categories above are 17, 52 and 87. Each step is the size of 35 since if a number has to both be evenly divisible by 5 and 7 it has to have 5 and 7 as factors, or 35 as a factor. So that's why the size of each step is 35.

(As shown: 17+35 = 52; 52+35 = 87)

 

But why does it start at 17?

If you'd write an expression for every possible x it would be "17 + 35n".

This is what I don't understand. Where does this 17 come from? Why is 17 the first number that fits the requirements for x? It must have something to do with the - 12 and + 4, but I don't know what. I only solved the problem by testing, not with a methodical approach.

plz help me

 

 

Answer 1

Find x such that for x chairs,
when they are placed in 5 rows there are 12 chairs left over
when they are placed in 7 rows there are 4 chairs too little

And that means:
    (x – 12) must be evenly divisible by 5;
    (x + 4) must be evenly divisible by 7;

The positive integers evenly divisible by 5 are 5, 10, 15, 20 ,,,
That means the possible x values are 17, 22, 27, 32, … [that is. 1*5+12, 2*5+12, 3*5+12, …]

The positive integers evenly divisible by 7 are 7, 14, 21, 28, …
That means the positive x values are 3, 10, 17, 24, … [that is, 1*7-4, 2*7-4, 3*7-4, …]

And the LCM of 5 and 7 is 5*7=35. That means that 35n is when this condition occurs (repeats). The lowest shared value is 17 (so let n=0 for this). That gives: x = 35n + 17 for n=0, 1, 2, 3, …

Calculate similarly for other size rows and other remainders.

Answer 2

Oliver, it took me some time to figure this out. Where does x and r come in if you have n in your expression??

 

You already answered your own question. 17 is the smallest number that fits both criteria.

 

5 rows with 1 chair and 12 extra chairs is 17 total chairs. 5 rows with 2 chairs and 12 extras gives us 22 total chairs. You just keep adding 5 because you add a chair to each of the 5 rows.

 

7 rows with one chair while missing 4 chairs gives us 3 chairs. That's just silly, because we only have 3 rows. If we add a chair to each of the 7 rows, we now have 10 chairs. Doing it again gives us 17 chairs.

 

Having 3 or 10 chairs does not meet the criteria for the first part, since it's not enough chairs to have 12 left over after making 5 rows.