Cubic display

Algebra Level 1

In a store display, identical cubic boxes are stacked in square layers. Each layer has 1 fewer row and column than the layer directly beneath it. If the bottom layer has 100 boxes and the top layer has 1 box, how many boxes are used in the display?

Image Credit: Flickr moggsterb

The answer is 385.

8 solutions

Angela Fajardo
Apr 8, 2015

1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 385

It is squared since they are stacked in square layers

If we look at it by layer, the number of boxes are: 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. And adding all of them:

1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 = 385

So there are 385 boxes used in the display.

Moderator note:

Good. Is there a systematic way to calculate the total number of boxes if the bottom layer has 10000 boxes instead?

You can use the formula

n(n + 1)(2n + 1)/6 , n = 10

Gamal Sultan - 6 years, 1 month ago

Brock Brown
Apr 12, 2015

Python 2.7:

from math import sqrt
def boxes(bottom):
    if not square(bottom):
        raise Exception, "Invalid bottom!"
    total = 0
    for layer in xrange(1,int(sqrt(bottom))+1):
        total += layer**2
    return total
def square(n):
    test = 0
    while test**2 <= n:
        if test**2 == n:
            return True
        test += 1
    return False
print boxes(100)

thats cool :-)

Arun Kumar - 6 years, 1 month ago

Ahmed Hazem
Apr 27, 2015

1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 385 you can even try it with legos :) , or any kind of cube...........

Karandeep Singh Ludhar
Apr 8, 2015

Since the boxes are displayed in square ways therefore there it will be in a form of 10 by 10 grid . First there are 100 boxes . Then 1 row and column is removed therefore their will be 81 boxes , then 64 , and so on . Therefore total boxes will be equal to the sum of square of first 10 natural numbers.

Do you know the formula for the sum of squares of first few positive integers?

Chung Kevin - 6 years, 2 months ago

I just know the formula that is n(n+1)(2n+1)/6, but can you please tell the proof , I am very confused about it .

karandeep singh ludhar - 6 years, 2 months ago

n^3 - (n + 1)^3 = -3 n^2 - 3 n - 1

1^3 - 2^3 = -3 (1^2) - 3 (1) - 1

2^3 - 3^3 = -3 (2^2) - 3 (2) - 1

3^3 - 4^3 = -3 (3^2) - 3 (3) - 1

4^3 - 5^3 = -3(4^2) - 3 (4) - 1

..........................................................

n^3 - (n + 1)^3 = -3 n^2 - 3 n - 1

Adding we get

1^3 - (n + 1)^3 = -3 (sum of squares of n natural numbers)

-3(sum of n natural numbers)

-(sum of n ones) ................................................. (1)

Let (sum of squares of n natural numbers) be denoted by S

(sum of n natural numbers) = n (n + 1)/2

(sum of n ones) = n

Substituting in (1) we get

-(n^3 + 3 n^2 + n) = -3 S - 3 n(n + 1)/2 - n

S = n(n + 1)(2 n + 1)/6

Gamal Sultan - 6 years, 1 month ago

@Gamal Sultan – Great proof! Better than induction since you don't have to know the formula in advance!

Otto Bretscher - 6 years, 1 month ago

Mathematical Induction.

Pi Han Goh - 6 years, 2 months ago

Raj Ayrus
Apr 28, 2015

1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + 7^2 + 8^2 + 9^2 + 10^2 = 385

Gamal Sultan
Apr 27, 2015

1^2 + 2^2 + 3^2 + .............. 10^2

= n(n + 1)(2 n + 1)/6 , n = 10

= 385

Jesus Frankenstein
Apr 27, 2015

question: how to solve this using diferential calculus?

or an approach to consider it the volume of a pyramid?

Any thoughts on how to answer your question?

Chung Kevin - 6 years, 1 month ago

Sha Man
Apr 27, 2015

1^2+2^2+3^2+....+10^2

Cubic display

Image Credit: Flickr moggsterb

The answer is 385.

8 solutions

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