House Of Cards

Algebra Level 2

Consecutive​ towers are built, as shown in the figure above.

The 1 st 1^\text{st} tower has one floor made of two cards.
The 2 nd 2^\text{nd} tower has two floors made of seven cards.
The 3 rd 3^\text{rd} tower has three floors made of fifteen cards, and so on.

How many cards will the 100 0 th 1000^\text{th} tower have?


The answer is 1500500.

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7 solutions

Hung Woei Neoh
Jun 7, 2016

Notice that for the n n th floor of the tower, you have:

2 n 2n slanted cards and ( n 1 ) (n-1) horizontal cards.

Eg. From the 3 rd 3^{\text{rd}} tower:

The 1 st 1^{\text{st}} row has 2 2 slanted cards and 0 0 horizontal cards

The 2 nd 2^{\text{nd}} row has 4 4 slanted cards and 1 1 horizontal card

The 3 rd 3^{\text{rd}} row has 6 6 slanted cards and 2 2 horizontal cards

Therefore, we can say that the number of cards in the n n th tower, c n c_n is given by:

c n = p = 1 n [ ( 2 p ) + ( p 1 ) ] = p = 1 n ( 3 p 1 ) c_n = \displaystyle \sum_{p=1}^n \left[(2p) + (p-1)\right] = \sum_{p=1}^n (3p-1)

Therefore, the number of cards in the 100 0 th 1000^{\text{th}} tower:

c 1000 = p = 1 1000 ( 3 p 1 ) = 3 p = 1 1000 p p = 1 1000 1 = 3 1000 ( 1001 ) 2 1000 = 3003 ( 500 ) 1000 = 1501500 1000 = 1500500 c_{1000} = \displaystyle \sum_{p=1}^{1000} (3p-1)\\ =\displaystyle 3 \sum_{p=1}^{1000} p - \sum_{p=1}^{1000} 1\\ =3\dfrac{1000(1001)}{2} - 1000\\ =3003(500) - 1000\\ =1501500 - 1000\\ =\boxed{1500500}

Nice observation (+1)

Ashish Menon - 5 years ago

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Thanks ¨ \ddot\smile

Hung Woei Neoh - 5 years ago
Ashish Menon
Jun 6, 2016

Relevant wiki: Arithmetic Progressions

In figure 1, there are 2 2 cards.
In figure 2, there are 7 7 cards.
In figure 3, there are 15 15 cards.
(For better proof):-
In figure 4, there will be 26 26 cards.
In figure 5, there will be 40 40 cards.



So, the number of cards in each case is increasing by the arithmetic progression 5 , 8 , 11 , 5,8,11,\cdots .
Now, the second figure has 5 5 cards more than the first term. The third figure has 8 8 more cards than second figure and so on. So, we see that the number of cards in the n th n^{\text{th}} figure exceeds the number of cards in the ( n 1 ) th {(n - 1)}^{\text{th}} by the ( n 1 ) th {(n - 1)}^{\text{th}} term of the above stated AP. So, the number of cards in the 1000 th {1000}^{\text{th}} figure exceeds that of the 999 th {999}^{\text{th}} figure by the 999 th {999}^{\text{th}} term of the AP.

So, the number of cards in the 1000 th {1000}^{\text{th}} is 2 + sum of first 999 terms of the AP = 2 + 999 2 ( 2 × 5 + ( 999 1 ) × 3 ) = 2 + 999 2 ( 10 + 998 × 3 ) = 2 + 999 2 ( 10 + 2994 ) = 2 + 999 2 × 3004 = 2 + ( 999 × 1502 ) = 2 + 1500498 = 1500500 2 + \text{sum of first 999 terms of the AP}\\ = 2 + \dfrac{999}{2} \left(2×5 + (999 - 1)×3\right)\\ = 2 + \dfrac{999}{2} \left(10 + 998×3\right)\\ = 2 + \dfrac{999}{2} \left(10 + 2994\right)\\ = 2 + \dfrac{999}{2} × 3004\\ = 2 + (999 × 1502)\\ = 2 + 1500498\\ = \color{#3D99F6}{\boxed{1500500}}

Nice solution! These numbers are known as the "second pentagonal" numbers.

Another equivalent definition for them is the sum of the first n n consecutive integers greater than n , n, for example:

2 2

3 + 4 3+4

4 + 5 + 6 4+5+6

etc.

and the general formula is n ( 3 n + 1 ) 2 . \dfrac{n(3n+1)}{2}.

Eli Ross Staff - 4 years, 10 months ago

Wow, using progressions to solve this!

Hung Woei Neoh - 5 years ago

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Hehe thanks! ʕ•ٹ•ʔ

Ashish Menon - 5 years ago

Did the same way. :D

Abhay Tiwari - 5 years ago

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:D ʕ•ٹ•ʔ (+1)

Ashish Menon - 5 years ago
Elvin Ding
Jun 21, 2016

1st 2nd 3rd 2 7 15 and so on... (a geometric sequence)

Double the terms by a factor of 2 to find more factors

4, 14, 30...

4: 2 2, 4 1 14: 2 7, 14 1 30: 3 10, 5 6, 15 2 (look for sequences in the factor pairs of the values.) 4: 4 1 14: 2 7 30: 3 10

Sequence of 1,2,3 is just "n" Sequence of 4,7,10 is just 3n+1 Find the value by multiplying both, since they are just factors that share sequences. n(3n+1)=3n^2+n But since we multiplied by 2 to find more factors, divide sequence by 2 3 n 2 + n 2 \frac{3n^2+n}{2} Plug 1000 into the sequence, and you get 1500500.

Jerry Xuan
Jan 14, 2020

1st tower=2 cards 2nd tower=7 cards 3rd tower=15 cards 4th tower=26 cards

The difference between 1st and 2nd tower=5 cards The difference between 2nd and 3rd tower=8 cards The difference between 3rd and 4th tower=11 cards

Each time, the difference of the cards will go up 3

Therefore the formula is the nth tower has 1.5n^2 + 0.5n cards

Therefore the 1000th tower has 1.5(1000^2) + 500=1500500

Bruce Brewer
Jun 15, 2019

Observe the number of cards coming from the teepees and the number of cards coming from the floors.

From the teepee pieces:

t(1) = 2

t(2) = 2+4

t(3) = 2+3+6

...

t(n) = 2(1+2+3+...+(n-2)+(n-1)+n) = 2( n ( n + 1 ) 2 \frac{n(n+1)}{2} ) = n(n+1)

From the floor pieces:

f(1) = 0

f(2) = 1

f(3) = 1+2

f(4) = 1+2+3

...

f(n) = (1+2+3+...+(n-2)+(n-1)) = ( n 1 ) ( n 1 + 1 ) 2 \frac{(n-1)(n-1+1)}{2} = n ( n 1 ) 2 \frac{n(n-1)}{2}

Total:

S(n) = t(n) + f(n) = n(n+1) + n ( n 1 ) 2 \frac{n(n-1)}{2} = n( 3 n + 1 2 \frac{3n+1}{2} )

S(1000) = 1000( 3001 2 \frac{3001}{2} ) = 1,500,500

Alternatively, you can think of the pyramids as rows of triangles with the bottom floor missing.

S(1) = 1x3 - 1

S(2) = 1x3 + 2x3 -2

S(3) = 1x3 + 2x3 + 3x3 - 3

...

S(n) = 3( 1 + 2 + 3 + ... + n ) - n = 3((n(n+1))/2) - n = n((3n + 1)/2), which is the same solution as above.

Bruce Brewer - 2 years ago
Freja Østerbøg
Aug 5, 2016

This is what I did:

First I saw a pattern

2 2* step number + ( + ( step number 1 ) + -1 ) + answer from last step

Then I wrote a program (In Python 3):

i=1 #Step number
x=0 #Number of cards
n=1001 #Max steps + 1
while i <= n:
    if i < n:
        x = 2*i + i-1 + x
        i = i + 1
    elif i == n:
        print(x)
        break

Result:

1500500 1500500

Sayan Das
Jun 21, 2016

Notice the pattern, 1st system- 2 cards, 2nd system 4+2+1 cards,3rd system 6+4+2+3 cards.Also note that 1,3,...... are pyramid numbers. The 1000th system consists of - (2+4+6+....to the 1000-th term)+1000(1000+1)÷2=1500500 cards

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