One side of a parking stall is defined by a straight stripe that consi

BrentGMATPrepNow Thank you for the solution. However, I have a doubt.
The question says that 'the length, in inches, of each unpainted section is an integer greater than 2'. According to your variables, this means 0.5x > 2 => x>4.

Now, in x(3n - 1) = (2)(7)(29)
If we consider x = 29, then 3n - 1 = 14 which implies n=5 (Option A)

Please let me know the mistake here.

Thank you!

A solution in which x = 29 breaks one of the conditions/

We're told that "...the length, in inches, of each unpainted section is an INTEGER greater than 2"
0.5x = the length of each unpainted section
So, if x = 29, then the length of each unpainted section = 14.5, which is not an integer.

Cheers,
Brent

We know the length of each unpainted section is an integer, so it's a lot easier to call that length "d" (rather than "d/2"). Then the length of a painted section is 2d. We have n painted sections, n-1 unpainted sections, and thus a total length of 2dn + (n-1)d = 3nd - d. This equals 203, so 3nd - d = 203, and factoring, d(3n - 1) = (7)(29). Since the factors on the left side are integers greater than 1, one of them must equal 7 and the other must equal 29 (since 7 and 29 are prime, there's no other possibility). Since 3n - 1 cannot equal 7 if n is an integer, 3n - 1 = 29, and n = 10.

Dear experts chetan2u IanStewart

If I make \(x\) the subject in the final expression I get \(x = \frac{203}{1.5n -0.5}\) and then since \(x\) has to be an integer and greater than 4, the denominator can either be 29 or 7. However, only 7 gives an integer value of 5 for \(n\) which is the option (A).

Please help with my mistake here.

OFFICIAL EXPLANATION

Because the length of an unpainted section is an integer, the length of a painted section (twice as long as an unpainted one) must be an even integer. So the part I've highlighted in red above is not correct: you will not get an even integer value for x if the denominator is 29 or 7. You'll only get an even integer when the denominator is something like 29/2 or 7/2. It's a bit of a confusing way to think through the problem, which is why I used a different choice of unknowns in my solution above.

What's a quick way to factor 406? I usually start by dividing by 2. So i get 203 * 2, but how do i determine whether 203 is prime or not? Appreciate if you could provide an approach that makes factoring / identifying primes much quicker..

Many thanks!

Great question.
There's no perfect way to determine whether 203 is prime or can be expressed as a product of two other numbers.
This is where knowledge of divisibility rules comes in handy.
If we know those divisibility rules, we see that 203 isn't the visible by 2, 3 or 5.
But what about 7?
I can see that 210 is divisible by 7 (210 = 7 x 30), and since 203 is 7 less than 210, I know that 203 is divisible by 7 (203 = 7 x 29)

One side of a parking stall is defined by a straight stripe that consists of n painted sections of equal length with an unpainted section 1/2 as long between each pair of consecutive painted sections. The total length of the stripe from the beginning of the first painted section to the end of the last painted section is 203 inches. If n is an integer and the length, in inches, of each unpainted section is an integer greater than 2, what is the value of n ?

A. 5
B. 9
C. 10
D. 14
E. 29

We can see that the stripe begins and ends with painted sections. So, the stripe is made up of an equal number of painted sections and unpainted sections + 1 painted section.

Each painted section is equal in length to two unpainted sections, each of which we can assign length u. So, the length of each painted section + unpainted section combination = 2u + u.

Thus, for the entire thing, we have (n - 1)(2u + u) + 2u = 203, So, (n - 1)(3u) + 2u = 203 where n and u must be integers.

3un - 3u + 2u = 203

3un - u = 203

(3n - 1)(u) = 203 where 3n - 1 and u are integers

This is a GMAT question, and we know that n and u are integers. Thus, we should have some nice numbers to deal with. So, a good next move is simply to factor 203.

We see that 203 is odd and can't be divided by 3 or 5. So, we can try 7 and find that 203 = 7 x 29.

So, u and 3n - 1 are 7 and 29 or 29 and 7 respectively.

Since n is an integer, only 29 can equal 3n - 1, since there's no integer n such that 3n - 1 = 7.

So, n must be 10.

The answer is (C).

Another way to do it is to see that there will always be a one less unpainted section than painted section and just try the answer choices with u as the length of an unpainted section.

(A) 5 -> 5(2u) + 4(u) = 203 -> 14U = 203 -> u = 14.5 -> Doesn't work because u has to be an integer.

(B) 9 -> 9(2u) + 8(u) = 203 -> 26u = 203 -> Doesn't work since 203/26 isn't an integer (since 260/26 = 10 and 260 - 203 = 57 isn't divisible by 26).

(C) 10 -> 10(2u) + 9(u) = 203 -> 29u = 203 -> u = 7 2u = 14. We got integer lengths. So, 10 works.

We could also check to make sure we did everything correctly.

10(2u) + 9(u) = 10(14) + 9(7) = 203 -> Perfect

The correct answer is (C).

For n painted sections, we have (n - 1) unpainted sections in between.
The length of each unpainted section is say 'a,' then length of each painted section is '2a.' So think of each painted section as if it is composed of two 'a's. Then we have (n - 1) unpainted a's and 2n painted a's.

Overall length of all sections = 203 = [(n-1) + 2n] * a = (3n - 1) * a

Since 203 = 7 * 29 or 1 * 203 (n and a are integers), and a must be greater than 1,
the only possibility is a = 7 because 29 is of the form (3n - 1).

If 3n - 1 = 29, then n = 10

Answer (C)

­The sentence "that consists of n painted sections of equal length with an unpainted section 1/2 as long between each pair of consecutive painted sections" is confused. 
I want to know why "between each pair of consecutive painted sections" could not be understood as below:

If it is, the approach will come up with another equation. I'm not a native speaker, please help.
 ­

The phrase 'between each pair of consecutive painted sections' means there is something between every two adjacent painted sections, indicating that there's something between each of the two sections. It does not mean that there is something between two pairs of sections.

Can someone please help me. I understand the previous explanations but the way I solved the question is giving me the wrong answer.

Here's how I attempted it.
Length of unpainted section = n
Length of painted section = 2n
Number of unpainted sections = k
Number of painted sections = k+1

Solving:
nk + 2n(k+1) = 203
nk + 2nk +2n =7x29
3nk + 2n =7x29
n(3k+2)=7x29

1) n=7 3k+2=29 => k=9
2) n =29 3k+2=7 => k=5/3

Why am I not getting the correct answer?
Why should it matter if I choose the number of painted sections as k+1 and unpainted as k instead of vice versa like the previous commenters have done it?

Bunuel

You are doing several things wrong there:

1. We are told that the number of painted sections is n, not that the length of a painted section is n. That will make the number of unpainted sections equal to n - 1.

2. The length of an unpainted section is 1/2 of the length of painted sections, so if you denote the length of unpainted sections as x, the length of painted sections will be 2x.

So, we get n(2x) + (n - 1)(x) = 203.