Maths Olympiad Prep

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Number theory Difficulty 3.9 AMC 10/12 Find the answer Canada

In the triangle shown, the first diagonal line, 1,2,3,4,1,2,3,4,\dots, begins at 1 and each number after the first is one larger than the previous number.

The second diagonal line, 2,4,6,8,2,4,6,8,\dots begins at 2 and each number after the first is two larger than the previous number. The nthn^{th} diagonal line begins at nn and each number after the first is nn larger than the previous number. In which horizontal row does the number 2016 first appear?

Pick one

Solution

Consider the diagonal lines that begin on the left edge of the triangle and move downward to the right.

The first number in the nthn^{th} diagonal line is nn, and it lies in the nthn^{th} horizontal row.

For example, the first number in the 3rd3^{rd} diagonal line (3,6,9,12,)({\bf3},6,9,12,\dots) is 3 and it lies in the 3rd3^{rd} horizontal row (3,4,3)({\bf3},4,3).

The second number in the nthn^{th} diagonal line is n+nn+n or 2n2n and it lies in the horizontal row numbered n+1n+1.

The third number in the nthn^{th} diagonal line is n+n+nn+n+n or 3n3n and it lies in the horizontal row numbered n+2n+2 (each number lies one row below the previous number in the diagonal line).

Following this pattern, the mthm^{th} number in the nthn^{th} diagonal line is equal to m×nm\times n and it lies in the horizontal row numbered n+(m1)n+(m-1).

The table below demonstrates this for n=3n=3, the 3rd3^{rd} diagonal line.

mm
mthm^{th} Diagonal Number
Horizontal Row Number

1
3
3

2
2(3)=62(3)=6
3+1=43+1=4

3
3(3)=93(3)=9
3+2=53+2=5

4
4(3)=124(3)=12
3+3=63+3=6

5
5(3)=155(3)=15
3+4=73+4=7



mm
m×nm\times n
3+(m1)3+(m-1)

The number 2016 lies in some diagonal line(s).

To determine which diagonal lines 2016 lies in, we express 2016 as a product m×nm\times n for positive integers mm and nn.

Further, if 2016=m×n2016=m\times n, then 2016 appears in the triangle in position mm in diagonal line nn, and lies in the horizontal row numbered n+m1n+m-1.

We want the horizontal row in which 2016 first appears, and so we must find positive integers mm and nn so that m×n=2016m\times n=2016 and n+mn+m (and therefore n+m1n+m-1) is as small as possible.

In the table below, we summarize the factor pairs (m,n)(m,n) of 2016 and the horizontal row number n+m1n+m-1 in which each occurrence of 2016 appears.

Factor Pair (m,n)(m,n)
Horizontal Row Number n+m1n+m-1

(1,2016)(1,2016)
2016

(2,1008)(2,1008)
1009

(3,672)(3,672)
674

(4,504)(4,504)
507

(6,336)(6,336)
341

(7,288)(7,288)
294

(8,252)(8,252)
259

(9,224)(9,224)
232

(12,168)(12,168)
179

(14,144)(14,144)
157

(16,126)(16,126)
141

(18,112)(18,112)
129

(21,96)(21,96)
116

(24,84)(24,84)
107

(28,72)(28,72)
99

(32,63)(32,63)
94

(36,56)(36,56)
91

(42,48)(42,48)
89

(Note: By recognizing that when m×n=2016m\times n=2016, the sum n+mn+m is minimized when the positive difference between mm and nn is minimized, we may shorten the work shown above.)

We have included all possible pairs (m,n)(m,n) so that m×n=2016m\times n=2016 in the table above.

We see that 2016 will appear in 18 different locations in the triangle.

However, the first appearance of 2016 occurs in the horizontal row numbered 89.

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