The Year in a Diagonal.

If a number $n$ can be factored as $x\cdot y$ then $n$ will appear in row $x+y-1$ . For example the number $6$ appears in row $1+6-1=6$ but also row $2+3-1=4$ .

$2018$ has only two factorizations, so appears in rows $1+2018-1=2018$ and $2+1009-1=\boxed{1010}$

The $n$ th contains precisely all numbers of the form $x\cdot y$ where $x + y = n+1$ . For instance, the 6th row contains $1\cdot 6 = 6,\ \ 2\cdot 5 = 10,\ \ 3\cdot 4 = 12,\ \ 4\cdot 3 = 12,\ \ 5\cdot 2 = 10,\ \ 6\cdot 1 = 6.$ To answer the question for a number $A$ , we must therefore find two factors $x\cdot y = A$ with $x + y = n + 1$ minimal. This minimum is obtained when $x,y$ are as close to each other as possible, and as close as possible to $\sqrt A$ ; if we list the factors of $A$ from smallest to greatest, they will be the two factors in the middle of the list, or the one factor $\sqrt A$ in the middle if $A$ is a perfect square.

The factors of $A = 2018$ are: $1,\boxed{2,1009},2018$ Thus $n+1 = 2 + 1009 = 1011$ and the answer is $n = \boxed{1010}$ .

Proof : If $x\cdot y = A$ , let $f(x) := x + y = x + A/x$ . Moreover, if we assume $x \leq y$ then $1 \leq x \leq \sqrt A$ . The derivative is $f'(x) = 1 - \frac{A}{x^2} \leq 0$ on the given domain, showing that $f$ is a decreasing function there. Thus the closer we choose $x$ to $\sqrt A$ , the smaller $n = f(x)$ will be.

The general expression for each row, where $r$ is the row number and $n$ is the term, is

$n(r-n+1)$

Solving for $r$ where this is equal to $2018$ , we get

$r=\dfrac{2018-n+n^2}{n}$

so that the mininum positive integer $r$ is $1010$ .

Each number is the product of the numbers at the tops of either diagonal that pass through it. For example on the above, $8=2\times 4$ and $8=4\times 2$ .

This means we can work out the row of a number by adding one of its cofactor pairs, then subtracting $1$ (as both diagonals share the starting 1). Using the same example, $8$ appears on the line $2+4-1=5$ .

The factors of $2018$ are $1$ , $2$ , $1009$ & $2018$ . This means the number $2018$ will appear on rows $2+1009-1=1010$ and $1+2018-1=2018$ , the lower of which is $\boxed{1010}$ .

If we watch the numbers from left to right the first numbers is always the rows number(rn), the second number(x) is twice then the previous row number. Inversely: rn=(x/2)+1 now x=2018.

rn=(2018/2)+1=1009+1=1010

Taking the only one of two factors of 2018...1009.

2018 will now be just below 1009th row... i.e 1010th row

We can simple divide 2018 by 2 and get 1009 and round it of to 1100 as this is a Pascal's triangle and two is the smallest factor of 2018 other than 1.

I just figured out that the first one it would show up in would be the one that would be around half of the end number as X 2 will be the earliest it can be seen other than in it's own row. Then by looking at the pattern I saw that the second number in each row was double the first number minus 2. From there, look at the options and 1010 X 2 = 2020 - 2 =2018. It will be the second number in the 1010th row.

Note that the second entry on each row is given by 2n -2 and are all multiples of 2. So if 2n - 2 = 2018, 2n = 2020, and n = 1010. Ed Gray

Deducing a formula by extending the series 7 12 15 16 15 12 7 8 14 18 20 20 18 14 8

R2 = x + (x-2) ; where R2 is the second element in the row from left and x is the row number. R2 = 2018; x = ?

Solving the equation 2x=2020; Thus x=1010

After I looked at this triangle I observed a simple pattern which only even numbers followed. To find the row in which they first appeared just divide the number by 2 and add 1 to the answer. For example :-
4 , 4/2 +1 row , 3rd row.
6 , 6/2 +1 row. , 4th row.
.....
2018, 2018/2 +1 row, 1010th row

First of all, the number 2018 cannot appear in the row whose value is the quotient of 2018 and 1,3,4 or 5. This is is shown by the answer choices. By this I mean 2018 divided by 1 is 2018 and so this number cannot be as it’s not in the list, and 2018 divided by 3, 4 or 5 does not give a whole number.

This division I’m doing may seem a little unclear. I’m dividing numbers because the each pattern is basically the 1, 2, 3, 4 or 5 times table, so dividing numbers finds how many times each number goes into 2018, and therefore the row number.

2 is the only number which goes into 2018; it goes 1009 times. But there’s no 1009th row you ask. Well the 2 times table doesn’t appear until the 2nd row, so you add one to 1009, giving the 1010th row.

Notice that each of the two sides increase by one, thus 2018 must appear on the 2018th row on either sides, but can it appear sooner? Well, what about the side that increases by two? with a simple division 2018/2 you get 1009, meaning that it will appear on the 1009th row (counting from the first row that has 2). Finally, you just add the first row that contains 1, and you get 1010th row (the right answer ).

Factors of 2018 are 2,1009 From symmetry, 3,5,....are prime numbers and their double i .e, (3×2)OR(5×2) appears at the the next of 3rd or 5th rows respectively. Such that ,2018 will appear to the next row of 1009 i .e, 1010 th row.

It is seen that every even no. (n) appears for the first time in the (n/2)+1 row. So, 2018 appears in the row: => (2018/2)+1 => 1009 + 1 = 1010

Each row is marked on the sides with its row number, and the sum of the two markers is always in the next row below. If 2018 is the result of two markers, that is 1009, and the row after that is the 1010th.