# Puzzle-like problem

• Jan 14th 2011, 04:43 PM
Hellbent
Puzzle-like problem
Hi,

On a square game-board that is divided into n rows of n squares each, k of these squares lie along the boundary of the game-board. Which of the following is a possible value of k?

A. 10
B. 25
C. 34
D. 42
E. 52

I have 25 (B). I don't have much reasoning behind my choice. Not sure if it requires much reasoning anyway. It mentions squares and the only number that is a perfect square is 25.
• Jan 14th 2011, 04:50 PM
Quote:

Originally Posted by Hellbent
Hi,

On a square game-board that is divided into n rows of n squares each, k of these squares lie along the boundary of the game-board. Which of the following is a possible value of k?

A. 10
B. 25
C. 34
D. 42
E. 52

I have 25 (B). I don't have much reasoning behind my choice. Not sure if it requires much reasoning anyway. It mentions squares and the only number that is a perfect square is 25.

The perimeter is k, which is 2n+2(n-2)=4n-4

10=14-4 but 14 is not a multiple of 4
25=29-4
34=38-4
42=46-4
52=56-4

56 is the only integer multiple of 4.
• Jan 14th 2011, 04:50 PM
Quacky
Edit: Nevermind, too late. Seriously, why do I even bother trying?(Headbang)(Punch)(Angry)
• Jan 14th 2011, 04:57 PM
Quote:

Originally Posted by Quacky
Edit: Nevermind, too late. Seriously, why do I even bother trying?(Headbang)(Punch)(Angry)

You may well have said it in a much more eloquent way!
Anyways... I was playing chess a couple hours ago
• Jan 14th 2011, 05:06 PM
Quacky
Quote:

You may well have said it in a much more eloquent way!
Anyways... I was playing chess a couple hours ago

Yup. Dirty cheat. *Incomprehensible comment about how my reasoning just complicated the question even further*
• Jan 14th 2011, 05:19 PM
Hellbent
Quote:

The perimeter is k, which is 2n+2(n-2)=4n-4

10=14-4 but 14 is not a multiple of 4
25=29-4
34=38-4
42=46-4
52=56-4

56 is the only integer multiple of 4.

Yeah, I never had faith in my approach. I don't follow this: 2n+2(n-2)=4n-4.

I think I follow the rest: a square has 4 sides, so you replaced n with each answer choice in the formula and checked which numbers would give a multiple of 4. Am I following this correctly?
• Jan 14th 2011, 05:22 PM
Quote:

Originally Posted by Hellbent
Yeah, I never had faith in my approach. I don't follow this: 2n+2(n-2)=4n-4.

I think I follow the rest: a square has 4 sides, so you replaced n with each answer choice in the formula and checked which numbers would give a multiple of 4. Am I following this correctly?

It's different to finding a perimeter.
We add all four side lengths there.

The problem here is that if you add up all the squares on the 2 horizontal sides, that's 2n.
In adding the squares left to be counted on the vertical sides,
notice that we've already accounted for the 4 squares at the corners!
So the amount of squares remaining to be counted is (n-2)+(n-2).
• Jan 14th 2011, 06:06 PM
Soroban
Hello, Hellbent!

Another way to count the perimeter . . .

Quote:

$\displaystyle \text{On an }n\times n\text{ chessboard, }k\text{ squares lie on the boundary of the board.}$
$\displaystyle \text}{Which of the following is a possible value of }k?$

. . $\displaystyle (A)\; 10 \qquad (B)\;25 \qquad (C)\;34 \qquad (D)\;42 \qquad (E)\;52$
Code:

       : - -  n-1  - - :       ♥ ♥ ♥ ♥ . . . ♥ ♥ ♠ -     - ♣                ♠ :     : ♣                ♠ :     : .                ♠ :     : .                .n-1   n-1.                . :     : ♣                . :     : ♣                ♠ :     : ♣                ♠ -     - ♣ ◊ ◊ . . . ◊ ◊ ◊ ◊         : - -  n-1  - - :

Each of the four sides has $\displaystyle n-1$ squares.

Hence, there are: .$\displaystyle 4(n-1)$ boundary squares, a multiple of 4.

The only multiple of 4 is: .$\displaystyle (E)\;52$

• Jan 15th 2011, 11:54 AM
DrSteve
Here is the way I would do this problem on the SAT:

First I draw some pictures.

**
**

***
***
***

****
****
****
****

Above are pictures for n=2,3,4
The corresponding values for k are as follows:

n=2 k=4
n=3 k=8
n=4 k=12

Draw as many values of n as you need to until you realize that the possible values of k are all positive multiples of 4.

Choice (E) is the only multiple of 4.