Trapezoidal Numbers

$T(x,y)=\frac{1}{2}(y-x+1)(y+x)$

which cannot be a power of $2$ , since both factors cannot both be even. Now, since the problem asks to find the "only number" between $10000$ and $20000$ , the number has to be $16384=2^{14}$ . Proving that it is the only number is another matter.

All the odd numbers can be expressed as $T(x,x+1) = 2x+1$ .

Let all even numbers with at least one odd factor be expressed as $(2a)(2b+1)$ . Then

$y=2a+b$
$x=2a-b$

or

$y=b+2a$
$x=b-2a+1$

whichever is greater between $2a, b$ .

The only numbers that cannot be expressed in this way are even numbers with no odd factors.

A slightly less formal way to see that any number $k \in \mathbb{N}$ with at least one odd factor (i.e. a number that's not a power of $2$ ) can be written as a trapezoidal number: Let $k = m(2n + 1)$ , for $m,n \in \mathbb{N}$ . Then if we use $m$ as the middle term in a series (strictly speaking, a summation), and then add in the $n$ integers directly before and after $m$ , the series will have a sum of $m(2n + 1)$ , because in a sequence consisting of an odd number of consecutive terms, the middle term is always the average.

For example, for $k = 77 = 7 \times 11$ , we can start with $7$ , and then add in the five integers directly before and after $8$ :

$77 = 2 + 3 + 4 + 5 + 6 + {\color{#D61F06}7} + 8 + 9 + 10 + 11 + 12 = T(2, 12)$

It is possible that there may be some negative integers in the series when first written; e.g. for $k = 52 = 4 \times 13$ , we start with $4$ , and then add in the six integers directly before and after $4$ , which gives us

$52 = \text{-}2 + \text{-}1 + 0 + 1 + 2 + 3 + {\color{#D61F06}4} + 5 + 6 + 7 + 8 + 9 + 10$

but then we can easily just remove the negatives, their positive counterparts, and zero, which will leave the same sum with a proper trapezoidal representation:

$52 = 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = T(3, 10)$

Note we can always do this 'removal', as there have to be fewer negatives than positives in the initial series, otherwise the initial $k$ would also be negative.

$T(x,y)\ = \frac { y(y+1) }{ 2 } -\frac { x(x-1) }{ 2 }\\ \quad =\frac { { y }^{ 2 }+y-{ x }^{ 2 }+x }{ 2 }\\ \quad = \frac { (y+x)(y-x+1) }{ 2 }$

Setting $y-x+1$ equal to $2$ :

$y-x+1=2\\ y+x = n\\ \rightarrow\ 2y-1=n$

Initially it appear that $n$ must be an odd number, but this is only because we set $y-x+1$ equal to 2. If we instead set it equal to 3:

$2y-2=n$

Therefore $n$ must have at least one odd factor, and as such for $n$ to not be a solution, it must be a power of 2. The only power of 2 within the range is $16384$ .

We can obtain a expression for $T(x,y)$ :

$\begin{aligned}T(x,y)&=x+(x+1)+\dots+y\\&=(1+2+...+y)-[1+2+...+(x-1)]\\&=\frac{y(y+1)}{2}-\frac{x(x-1)}{2}\end{aligned}$

And let an integer $n$ be its value. Then $\begin{aligned}2n&=y^2+y-x^2+x\\&=(y+x)(y-x)+(y+x)\\&=(y+x)(y-x+1)\end{aligned}$ Now we split in two cases.

When $x\equiv y\pmod2$ , we have $n=\dfrac{y+x}2(y-x+1)$ . Because $y>x$ , the term $y-x+1$ must be an odd number larger than 1.

When $x\not\equiv y\pmod2$ , we have $n=(y+x)\dfrac{y-x+1}{2}$ . Because both $x,y$ are positive integers, so $y+x$ would also be an odd integer larger than 1.

As a result, we cannot have $n$ to have the only factor $2$ , so the only answer cannot obtain is $2^{14}=\boxed{16384}$ .

We can produce any odd number $n$ with $T(\frac{n-1}{2},\frac{n+1}{2})$ .

We can produce any multiple of any odd number, as the number of circles in an odd-height trapezium is equal to its height multiplied by length of the middle row.

This means we must find an even number with only even prime factors. As $2$ is the only even prime, our solution must be the only power of $2$ between $10,000$ and $20,000$ : $2^{14}=\boxed{16384}$ .

Generalization

Let the prime factorization of $N$ be $N = 2^{e_2}\cdot 3^{e_3}\cdot 5^{e_5}\cdots p^{e_p}\cdots.$ Then the number of ways in which $N$ can be expressed in the desired form is $X(n) := \#\{(x,y)|T(x,y) = N\} = (e_3+1)\cdot (e_5+1)\cdots (e_p+1)\cdots - 1.$ We see that $X(n) = 0$ iff $e_3 = e_5 = \cdots = 0$ , i.e. the prime factorization of $N$ no odd prime factors. Therefore $N$ is at best a power of 2, which in this case requires $N = \boxed{16384}$ .

Proof of this expression:

We have $N = T(x,y) = \tfrac12 y (y+1) - \tfrac12 x(x-1) = \tfrac12(y - x + 1)(y + x) =: \tfrac12 A B$ The factors $A = y - x + 1$ and $B = y + x$ have opposite parity, i.e. one is odd and the other is even.

Thus we are looking for factorings of the form $2N = AB$ , with $A$ and $B$ of opposite parity and $A<B$ . We take out all factors 2, i.e. we write $N = 2^{e_2} \cdot m\ \ \ \ m = 3^{e_3}\cdot 5^{e_5}\cdots p^{e_p}\cdots \ \text{is odd}.$ Let $m = uv$ be any factoring of $m$ ; now set $A = u$ and $B = 2^{e_2+1}v$ , and swap them if necessary to ensure that $A < B$ . Let $x = \tfrac12(B - A + 1),\ \ \ y = \tfrac12(B + A - 1).$ Then $T(x,y) = N$ as required; and the number of pairs $(x,y)$ is equal to the number of factors of $m$ , that is, $(e_3+1)\cdot (e_5+1)\cdots (e_p+1)\cdots$ .

However , for each value of $N$ we must rule out the trivial factoring with $u = 1$ . This leads to $A = 1$ and $B = 2N$ , so that $x = y = 2N$ , violating the condition $x < y$ . Therefore we subtract one and obtain the final count $X(n) = (e_3+1)\cdot (e_5+1)\cdots (e_p+1)\cdots - 1.$

To illustrate, consider the case $N = 90$ , with $e_2 = 1$ and $m = 45 = 3^2\cdot 5^1$ ; there are $(2+1)(1+1) = 6$ ways to factor $45 = uv$ :

$\begin{array}{cc|cc|cc} u & v & A & B & x & y \\ \hline 1 & 45 & 1 & 180 & 90 & 90 \\ 3 & 15 & 3 & 60 & 29 & 31 \\ 5 & 9 & 5 & 36 & 16 & 20 \\ 9 & 5 & 9 & 20 & 6 & 14 \\ 15 & 3 & 12 & 15 & 2 & 13 \\ 45 & 1 & 4 & 45 & 21 & 24 \\ \hline \end{array}$ The first row of this table violates the condition $x < y$ , so that we are left with $X(90) = 6 - 1 = 5$ .

With the help of some python code and a formula $T(a, b)=\frac{1}{2}(b+a)(b-a+1)$ :

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# Set of all integers 10000 to 20000
nleft = set(range(10000, 20001))

# Iterate through possible pairs (a, b)
for b in range(1, 10000):
    for a in range(1, b):
        # Apply formula
        r = (b + a) * (b - a + 1) // 2
        # Remove resulting integer from the set
        if r in nleft:
            nleft.remove(r)
    # Show progress
    print(len(nleft))
    if(len(nleft) == 1):
        break

# Print the set
print(nleft)

# Output
# nleft = {16384}

It is the power of 2 between 10000 and 20000, which is $\fbox{16384}$ .

See Rectangular and Trapezoidal Arrangements , where I show that the number of trapezoidal arrangements of $n$ items (including the degenerate case of all items on a line) equals the number of odd divisors of $n$ . The powers of 2 are the only numbers with exactly one odd divisor, viz. 1.

I struggled with this question and finally solved it. I found that this is level 2 in number theory, but I want to ask the experts how hard is this question

In python i created a list of numbers from 10000 to 20000, then i started writing functions to remove numbers based on rules for each stack: So we remove: 1 stack: not allowed 2 stacks: all numbers when added to 1 is divisible by 2 3 stacks: all numbers divisible by 3 4 stacks: all numbers when added to 2 is divisible by 4 5 stacks: all numbers divisible by 5 6 stacks: all numbers when added to 3 is divisible by 6

And you quickly realized the pattern for N stacks: If N is odd, all numbers divisible by N should be removed. If N is even, all numbers, when added to N/2, is divisible by N should be removed. So just write one single function with these two rules and run the list through it. And voila!

This method is a bit inductive but it does help here:

T(x,y)= =z(say)=x+(x+1)+(x+2)+…+[x+(y-x)]

=((y(y+1))/2)-((x(x-1))/2)

=(y+x)(y-x+1)/2

Now, y>x. So, let y=x+n.

z=(2x+n)(n+1)/2.

Let $Z_n$ represent the set of values of z corresponding to n.

Now, $z_k$ = $(k+1) \times ((2x+k)/2)$

Case 1: k is odd :

k+1 is even.

Then, k+1/2 may be odd or even.

But, $2x + k$ is odd ( $even + odd$ = odd).

So, $z_k$ = $odd \times odd$

Or, $z_k$ = $odd \times even$

Case 2: k is even :

$2x + k$ is even ( $even + even$ =even) But k+1 is odd.

So, $z_k$ = $odd \times even$

Reader may verify that $Z_1$ ={3,5,7,9,11,…} covers all odd numbers except 1.

Let's hunt for even numbers.

The even numbers are of the form $odd \times even$ , which rules out the presence of (integral) powers of 2 in any of the set $Z_k$ .

The only (integral) power of 2 between 10000 and 20000 is 16384.

Hence, the answer is 16384.

Another way of interpreting the set of all possible values of $T(x, y)$ is as the set of all possible sums of two or more consecutive positive integers. Now, any number with an odd factor is expressible as the sum of consecutive integers. For example, if $n = 5*k$ , then $n = (k-2) + (k-1) + k + (k+1) + (k+2)$ . The only number between 10000 and 20000 that doesn't have an odd factor is $2^{14} = 16384$ .

Let us express T as a function of n and y where n is the number of slabs and y is the number of balls in the bottom row. So, T(n,y) = y + (y-1) + (y-2) + ...... + (y-n+1) T(n,y) = ny - (1 + 2 +3 + ..... + (n-1)) T(n,y) = ny - $\frac{n(n-1)}{2}$
Now, letting T(n,y) = k and rearranging the expression, we get y = $\frac{k}{n}$ + $\frac{(n-1)}{2}$ .If n is odd, for y to be integer, k has to be a multiple of n .So k can be of the form 3n , 5n , 7n .... (n cannot be 1) .If n is even, for y to be integer, k has to be an odd multiple of $\frac{n}{2}$ .So, k should be of the form 2n+1 , 4n+2 , 6n+3 , 8n+4 ....... .So, all odd numbers can be achieved as they are of the form 2n+1 .This leaves us with even numbers .All even numbers are either of the form 4n or 4n+2 .4n+2 can be achieved .So, we are left with 2 and multiples of 4 .2 because in 4n+2, n cannot be 0 .All multiple of 4 can be expressed as 8n or 8n+4 .8n+4 can be achieved .So, we are left with 2,4 and multiples of 8 .This pattern tells us that a number of the form 2^x cannot be achieved if n is even .This form cannot be achieved if n is odd .So the question reduces to finding the number of the form 2^x that lies between 10000 and 20000 .The answer is 16384 (2^14)

Let us think about sorting all possible values of $T(x,y)$ where $x<y$ by the number of rows.

As $x<y$ , we have 2 rows at least. In the situation of 2 rows, the minimum possible value is $T(1,2)=3$ , and all other values can be represented as $3+2k,$ where $k$ is a non-negative integer.

Similarly, in the situation of $n$ rows, the possible values are $T(1,n)+nk,$ where $k$ is a non-negative integer.

For $T(1,n)=\dfrac{n(n+1)}{2}$ , we can make a statement as follow. If $n$ is even, then $T(1,n)=\frac{n(n+1)}{2}\equiv \frac n2({\rm mod}\,n).$

If $n$ is odd and $n\ne 1$ , then $T(1,n)=\frac{n(n+1)}{2}\equiv 0({\rm mod}\,n).$

Now consider the situations where $n=2,4,8,\ldots,2^m\ldots$ . These situations tell us that the values in the form $2k+1, 4k+2, 8k+4, 16k+8, \ldots, 2^mk+2^{m-1}, \ldots$ are all possible. And they can quickly cover almost every positive integer, except those of the form $2^m$ .

And between 10000 and 20000 there is only one single number is the power of 2. It is 16384.

On the other hand, if $n$ is odd and $n\ne 1$ , there is no chance for $n$ to become a factor of 16384.

And If $n$ is an even number but not a power of 2, $T(1,n)$ at least has an odd factor which is not 1. Therefore 16384 is not a possible value here.

In conclusion, 16384 is the ONLY number between 10000 and 20000, who can't be a value of $T(x,y)$ , where (x<y).