Pyramid problem

Question

If you have 140 stone blocks, you can build an Egyptian square pyramid whose base has 7 blocks on each side. At the same time, you can arrange those blocks into 35 perfect 2 × 2 squares. Find a positive integer value of n>100 for which 35 perfect n*n squares can also be arranged into an Egyptian square pyramid.

Answer 1

1. It turns out that the equation for square pyramidal numbers is P = [n* (n+1)*(2n+1)] / 6  {Eqn 1}.  Knowing this formula is obviously one key to solving the problem.
2. The formula above will tell you all the numbers P that form square pyramidal numbers.  For example, when n = 2, the number of blocks required is (2*3*5)/6 = 5 = P.  As another example, we can use the information given in the first couple of sentences of the problem.  Here n =7, and P = (7*8*15) = 140.  Also note that these 140 blocks can be arranged in 35 perfect 2x2 squares because a 2x2 square would require 4 blocks, and 140 total blocks / 4 blocks in a group = 35 groups.
3. This problem is just asking to find a value P = 35(X*X) → P = 35X²
4. So substituting into Eqn 1:  35X² = [n* (n+1)*(2n+1)] / 6
5. Divide both sides by 35:  X² = [n* (n+1)*(2n+1)] / 210
6. Take the square root of both sides:  X = sqrt([n* (n+1)*(2n+1)] / 210)    {Eqn 2}
7. Now, as far as I can tell, this becomes less of a math problem and more of a "hunting" game.  The "hunt" requires that you pick a number n such that the radicand on the Right Hand Side of Eqn 2 is a perfect square.
8. So what do we know?  We know that the term under the radical ([n* (n+1)*(2n+1)] / 210) must be a perfect square.  Why?  Because you will be taking the square root of this number, and the result must be an integer.  Why?  Because you can't have a fractional number of blocks;  you must have an integer number of blocks
9. Given that the radicand is an integer, what else do we know?  We know that if the radicand to be an integer, the numerator must be a multiple of the denominator 210.  So, for example, the numerator can be 2*210 = 420.  When we divide 420/210, we get an integer 2.
10. So now we simply are hunting for a number X, which will make the numerator a multiple of 210, AND which results in the radicand being a perfect square.   One efficient way to do this would be to create a spreadsheet.  I don't know if using a spreadsheet is an allowable method in your particular case.  But I developed a spreadsheet that simply goes thru the numbers (n)  1 thru 1000 (I decided to stop at 1000, but there was no guarantee that I would find a solution and so I was ready to increase the number to 10000) and looks for a corresponding X value that is an integer.  So I had one column that was n, and the rows went from 1 to 1000.   The first time I see a value of X which is an integer, then I have found my answer.
11. Having done this, I discovered that when n = 840, the value of X is an integer (2378).  This is the first number X>100 for which the value of X is an integer.
12. So the answer to the question is X = 2378. (Sorry, I realize now that the original problem was calling this variable "n", but I used n differently as given by Eqn 1.  Hope this isn't confusing.  Just assume that the original problem statement is "Find a positive integer value of X>100 for which 35 perfect X*X Squares can also be arranged into an Egyptian square pyramid")
13. Let's just interpret what these results mean.  The value n of 840 means there are 840 levels in the pyramid.  The total number of blocks P required is given by Eqn 1, and is equal to 197,920,940.  The problem asked for being able to arrange this many blocks in 35 groups of X² blocks.  If we square 2378, we get 5,654,884.  So we can create 35 groups of blocks, where each group has 5,654,664 blocks.  Each group can be arranged as 2378 blocks x 2378 blocks.
14. If you are not allowed to use a spreadsheet to solve this problem, then you would need to go through all the numbers beginning with n = 101 until you find a perfect square.  That would take a very, very long time.  So perhaps there is some trick to quickly finding perfect squares that I am not familiar with. Or maybe there is a better way to solve this problem.   I will think about it a little more, and update my answer accordingly.

 

***********************  UPDATE ******************

I had a little bit of a revelation after looking at this problem again this evening, when I thought that it might be more than a coincidence that the n value of 840 happened to be a multiple of the denominator (210).  The key is that the problem doesn't require finding the FIRST value of X > 100 where 35 perfect X * X squares can be arranged.  For some reason, I thought it was important to find the first occurrence.  So, the question becomes whether there is some simplification that occurs if we are looking for ANY value of X that satisfies the criteria.  And the answer is "yes".  So, when we get to step 9 above, instead of relying on a spreadsheet, here is how to proceed:

 

1. We're looking for ANY value of X that satisfies the criteria that  ([n* (n+1)*(2n+1)] / 210) must be a perfect square. The numerator must be a multiple of 210.  This isn't sufficient to find the answer, but it is necessary.  So, the simplification that we can make, since we're looking for ANY value of X, is to say that n must be a multiple of 210.  Why?  Because we know that when n is a multiple of 210, it will result in an integer number when it is divided by the denominator of 210.  Now we just need the other two terms (i.e., "n+1" and "2n+1") to create a situation where the three terms together create a perfect square.  If this isn't clear, the following steps may make it more obvious.  
2. Let's start by setting n = 210.  In this case, X = 283.302, which is not an integer and therefore cannot be the solution.  Also, note that I'm rounding the value of X.  What's important is whether it is an integer or not.  The accuracy is not important.  Move to 2*210.  With n = 420, X = 841.50.  Again, X is not an integer so can't be the answer.  Move on to 3* 210.  With n = 630, X = 1545.016.  Not an integer.  Move on to 4*210.  With n = 840, X = 2378.  Finally we have an integer. 
3. Notice that the answer of X = 2378 matches the solution we found in Step 12 above.  This is telling us that the X=2378 is not just ANY value of X that solves the problem, but in fact it is the FIRST value of X that does.
4. I think we just happen to be very lucky that it took only four attempts before we found the solution.  There is nothing that guarantees (as far as I can tell) that this method will give us a relatively quick solution for all variations of this problem.  For example, if the problem had stated 36 perfect 2 x 2 squares (instead of 35), then the method used here would not have worked in 4 attempts.  So, I think for the general case for this type of problem, a spreadsheet (or alternately computer program) is necessary to find a solution.

 

The bottom line is that, once we get past step 8,  neither the method used in my original response or the updated method is really using math properties to find the solution.  In both cases, we're just "hunting" for the right answer, as opposed to using some math principals to get the answer.   I would be very interested in knowing if someone has some other approach that relies on solving math equations, algebraically or otherwise,  to get the right answer.