Track / Stage 3 / 28 of 260 #508 of 2444

Problem 508

AMC 10/12, early questions

Combinatorics Difficulty 3.1 Prove it CEMC Galois · Canada · 2013

The even positive integers are listed in order and arranged into rows, as shown, and described below.

Each new row includes one more integer than the previous row. The last number in each row is the product of the row number and the next largest integer. For example, the last number in the 4 $^{th}$ row is $4\times 5$ . You may use this fact without proving it.

List the numbers in the 7th row of the table.
What are the first and last numbers in the 100th row of the table?
The last number in row $r$ is $L$ . The first number in row $(r+2)$ is $F$ . Determine the smallest possible value for $r$ such that $F+L$ is at least 2013.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution 1

The last number in the $7^{th}$ row is $7\times8=56$ .

Since the $7^{th}$ row has 7 numbers in it, we list the 7 even integers decreasing from 56, which are $56,54,52,50,48,46,44$ .

Written in the order they will appear in the table, the numbers in the $7^{th}$ row are, $44,46,48,50,52,54,56$ .

Solution 2

The last number in the $6^{th}$ row is $6\times7=42$ .

Therefore the next even integer, 44, will appear as the first number in the $7^{th}$ row of the table.

Since the $7^{th}$ row has 7 numbers in it, we list the 7 even integers increasing from 44.

Thus the numbers in the $7^{th}$ row are, $44,46,48,50,52,54,56$ .
The last number in the $100^{th}$ row is $100\times101=10\,100$ .

The last number in the $99^{th}$ row is $99\times100=9900$ .

Therefore the next even integer, 9902, will appear as the first number in the $100^{th}$ row of the table.

The first and last numbers in the $100^{th}$ row of the table are 9902 and $10\,100$ , respectively.
The last number in row $r$ is equal to $r(r+1)$ , so $L=r(r+1)$ .

The first number in row $(r+2)$ is 2 more than the last number in row $(r+1)$ .

The last number in row $(r+1)$ is $(r+1)(r+2)$ , so $F=(r+1)(r+2)+2$ .

We require $F+L$ to be at least 2013, so $F+L=(r+1)(r+2)+2+r(r+1)\geq2013$ .

To determine the smallest value for $r$ such that $F+L=(r+1)(r+2)+2+r(r+1)\geq2013$ , we solve the following inequality: $\text{(r+1)(r+2)+2+r(r+1) 2013 r 2+3r+2+2+r 2+r 2013 2r 2+4r+4 2013 r 2+2r+2 1006.5 r 2+2r+1 1006.5-1 (r+1) 2 1005.5 r+1 + 1005.5 or r+1 - 1005.5}$ Since $r$ is positive, $r+1\geq\sqrt{1005.5}$ and so $r\geq+\sqrt{1005.5}-1\approx30.7096$ .

Thus, the smallest possible value of the integer $r$ such that $F+L$ is at least 2013 is 31.

Check: Since $L$ is the last number in row $r=31$ , then $L=31\times32=992$ .

Since $F$ is the first number in row $r+2=33$ , then $F$ is 2 more than the last number in row 32, or $F=(32\times33)+2=1058$ .

Therefore, $F+L=1058+992=2050\geq2013$ as required.

We must also check if $31$ is the smallest value of $r$ such that $F+L\geq2013$ .

Since the numbers are arranged in the rows in a strictly increasing way, we need only check that when $r=30$ , $F+L<2013$ .

When $r=30$ , $F+L=((31\times32)+2)+(30\times31)=994+930=1924<2013$ .

Source: CEMC, University of Waterloo, licensed CC-BY-NC-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.