Question

It is desired to pile some logs in layers so that the top layer contains one log, the next layer contains two logs, the next layer contains three logs, and so on; each layer containing one more log than the layer on top of it. There are 190 logs. Can all the log be used in such a grouping, and if so, how many logs are in the bottom layer?​

Answer 1

Answer:

Yes all of the logs can be used in such a grouping.

We'll have 19 logs in the bottom row (and have 19 rows)

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Explanation:

Let n be the number of layers. Let T(n) represent how many logs there are total.

• If you have 1 layer, then you have 1 log. So we can say that T(1) = 1.
• If there are n = 2 layers, then we have 1 log on top of two other logs, which leads to 1+2 = 3 logs total. This shows that T(2) = 3.
• If there are n = 3 layers, then we have 1 log on top of 2 which are on top of 3 at the very bottom. We have 1+2+3 = 6 logs total. So T(3) = 6.
• Having n = 4 layers gives 1+2+3+4 = 10 logs. So T(4) = 10
• Having n = 5 layers gives 1+2+3+4+5 = 15 logs. So T(5) = 15

This process continues for however long you like. The sequence 1,3,6,10,15 is the set of triangular numbers, since the logs form a triangular or pyramid shape when stacked this way. To get the nth value in the triangular number sequence, use this formula

T(n) = 0.5*n*(n+1)

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For example, let's say we wanted to find how many logs there were in the fifth layer

T(n) = 0.5*n*(n+1)

T(5) = 0.5*5*(5+1)

T(5) = 15

This matches from earlier.

The purpose of bringing in a formula is to see if we can get T(n) = 190 for some integer n.

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T(n) = 0.5*n*(n+1)

190 = 0.5*n*(n+1)

190*2 = n(n+1)

380 = n^2+n

0 = n^2+n-380

n^2+n-180 = 0

Applying the quadratic formula will lead to the two solutions: n = -20 and n = 19. We ignore the negative value of n as we can't have a negative number of rows. Therefore, n = 19 is the only valid solution here.

If we wanted to stack 190 logs in this triangular pyramid fashion, then we'll need to have 19 rows. The bottom row will have 19 logs.

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Check:

1+2+3+4+5+6+7+8+9+10 = 55

11+12+13+14+15+16+17+18+19 = 135

55+135 = 190

Or you could say,

T(n) = 0.5*n*(n+1)

T(19) = 0.5*19*(19+1)

T(19) = 190

Either way, the answer is confirmed.