$A+B+C=2149$, Find $A$ In the following form of odd numbers 

If the numbers 
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
 A: If $A$ were on the $n^{th}$ row, then $B=A+2n$ and $C=A+2(n+1)$.
Out of laziness, we will try to figure out what row this would need to be to be in the right ballpark through estimation.  $\frac{2149}{3}\approx 716$, so $A<716<B<C$.  Which row would $715$ fit into?
By changing your figure some by first adding one to each entry and then dividing by two, we will get the chart $\left[\begin{array}{}\color{red}{1}\\2&\color{red}{3}\\4&5&\color{red}{6}\\7&8&9&\color{red}{10}\\\vdots\end{array}\right]$, and these numbers are very familiar to us.  The red numbers are the triangle numbers, $T(n)=\binom{n+1}{2}=\frac{n^2+n}{2}$.
So, $T(26)=351<\frac{715+1}{2}=358<378=T(27)$, so we suppose that $715$ occurs on the $26^{th}$ row, implying that either $A$ is on the $25^{th}$, $26^{th}$, or $27^{th}$ row (since we have been estimating up to this point).
So, we try to solve now, $A+B+C=2149=A+(A+2n)+(A+2n+2)=3A+4n+2$
In the case that $n=25$, this would be $3A+100+2=2149\Rightarrow 3A=2047\Rightarrow A=\frac{2047}{3}\not\in\mathbb{Z}$, so we know that $n=25$ was not possible.
In the case that $n=26$, this would be $3A+104+2=2149\Rightarrow 3A=2043\Rightarrow A=681$
In the case that $n=27$, this would be $3A+108+2=2149\Rightarrow 3A=2039\Rightarrow A=\frac{2039}{3}\not\in\mathbb{Z}$, so we know that $n=27$ was not possible.
We can similarly show that if $n=24$ that leads to a contradiction as well.
We believe then that our answer is $A=681, B=733, C=735$.  All that remains to check is that $681$ is indeed on the $26^{th}$ row to confirm our calculations by checking that $T(25)$ is less than $\frac{A+1}{2}$ and $T(26)$ is greater than $\frac{A+1}{2}$.
Indeed, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$
A: $A$ is going to be the $k$th entry in the $n$th row, where $1 \leq k \leq n$. That will make $B$ the $k$th entry in the $n+1$st row, and $C$ the $k+1$st entry in the $n+1$st row.
So the first question is, can we write a formula for $A$ in terms of $k$ and $n$? If we consider the first entry in the $n$th row, there are $1 + 2 + \cdots + (n-1)$ odd numbers before it, which equals $\frac{n(n-1)}{2}$. So the first entry in the $n$th row must be $2\left(\frac{n(n-1)}{2}\right) + 1 = n^2 - n + 1$. Now going over to the $k$th entry will add $2(k - 1)$ to this, so $A = n^2 - n + 1 + 2(k-1) = n^2 - n + 2k - 1$
This formula also implies that $B = (n+1)^2 - (n+1) + 2k - 1$ and $C = (n+1)^2 - (n+1) + 2(k+1) - 1$.
If we put this all together, $A + B + C = (n^2 - n + 2k - 1) + (n^2 + n + 2k - 1) + (n^2 + n + 2k + 1) = 3n^2 + n + 6k - 1$.
So can we solve:
$$3n^2 + n + 6k - 1 = 2149$$
Or:
$$3n^2 + n + 6k = 2150$$
Solving for $k$:
$$k = \frac{2150 - 3n^2 - n}{6}$$
Now we need $k$ to be an integer, and $1 \leq k \leq n$. Since $k$ is an integer we know $n$ should be $2$ mod 3, and we know $3n^2 \leq 2150$, so $n^2 \leq 717$. So $n < 27$. Now we guess, starting with the highest value of $n$ which is less than $27$ and $2$ mod 3, so $n = 26$. Plugging this in we get $k = 16$, which works!
This gives $A = 26^2 - 26 + 32 - 1 = 681$.
A: I like this one.
Look, let's fill the blanks in the middle column with pair numbers. You will find that it forms the sequence 1, 4, 9, 16, 25, 36.. that is evidently x².
All the numbers in a row are sequential, so we can establish a variable to this position i.e. n. And establish that in the A position we have n, so in the B position we have (n - 1) and in the C position we have (n + 1).
Given all this we can say:
A = x² + n
B = (x + 1)² + (n - 1)
C = (x + 1)² + (n + 1)
So we can put this in the original equation and we will have an equation with two values to solve (x, n). Nevertheless we also know that n can not be greater than x, ensuring the selected values are in the triangle. 
So
(x² + n) + (x + 1)² + (n - 1) + (x + 1)² + (n + 1) = 2149
3x² + 4x + (3n) + 2 = 2149
3x² + 4x + (3n - 2147) = 0
With this equation we can find:
x = 26
n = 5
And so:
A = 681
B = 733
C = 735
Hope it helps!
A: Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*}
&& 2149 &= a_k + 2a_{2k+n+1} +2 \\
&\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1
\end{align*}
Note that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where
$$k_n = 1+\cdots+n = \frac{n(n+1)}2.$$
Therefore, $k=k_n+q$ where $0\le q\le n$. Substituting, we get
\begin{align*}
&& 2147 &= 3n(n+1) + 10q + 4n + 1 \\
&\Rightarrow &  0 &= 3 n^2 + 7 n + 10 q -2146
\end{align*}
Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.
Indeed, plugging in $n=25$ into our earlier equation gives
\begin{align*}
&& 2147 &= 6*k + 4*25 + 1 \\
&\Rightarrow &  2046 &= 6*k \\
&\Rightarrow &  341 &= k
\end{align*}
So we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.
A: Looking at the first element in each row, we see
 row      A    B    C   A+B+C
   1      1    3    5       9
   2      3    7    9      19
   3      7   13   15      35
   4     13   21   23      57
   5     21   31   33      85
   6     31   43   45     139

Using finite differences, we find that 
\begin{align}
   A &= n^2 - n + 1\\
   B &= n^2 + n + 1 = A+2n\\
   C &= n^2 + n + 3 = A+2n+2\\
A+B+C &= 3n^2+n+5
\end{align}
where A is the first element in row $n$.
For each shift of A to the right in the same row, A,B, and C increase by 2
and A+B+C increases by $6$.
Solving $3n^2+n+5 = 2149$ for n, we get $n = 26.567$ So we are in row $26$.
For the first element in row $26,
 A = 651,\; B = 703,\; C = 705,$ and $A+B+C = 2059$.
Since $\dfrac{2149-2059}6 = 15$, we need to increase $A, B, $ and $C$ by $2\cdot15 = 30$.
So
\begin{align}
   A &= 681\\
   B &= 733\\
   C &= 735
\end{align}
A: Let $r$ be the row number, and $U$ be the unshown number between $B$ and $C$ and directly below $A$, i.e. 

$$\begin{align}
&A 
&&\leftarrow\text {row}\; r\\
B \quad [&U]\quad  C
&&\leftarrow\text {row}\; r+1
\end{align}$$

Note that the difference between two vertically adjacent numbers (both shown and unshown)   is $(r+1)^2-r^2=2r+1$, i.e. $U-A=2r+1$. Hence
$$\begin{align}
2149&=A+B+C&&\text{(given)}\\
&=A+2U&&\text{(as}\; B=U-1, C=U+1\text{)}\\
&=3A+4r+2&&\text{(using } U=A+2r+1\text{)}\\
3A+4r-2147&=0&& && .....(1)\\
\end{align}$$
Note also that the centre column (including unshown numbers) corresponds to $r^2$. As an initial approximation, assume that $A$ is the centre column, i.e. $A=r^2$. Substituting in $(1)$ gives 
$$\begin{align}3r^2+4r-2147&=0\\
\color{blue}{r}&\color{blue}{\approx 26} \; (r>0)\qquad \qquad &&  && && &&& .....(2)\end{align}$$
Putting $(2)$ in $(1)$ gives 
$$\color{red}{A=681}\qquad\blacksquare\qquad
\Rightarrow \color{blue}{B=733, C=735}$$
Check: $A+B+C=681+733+735=2149$
NB: The number "$A$" is in row $26$ but is not in the centre column; the element in the centre column is $26^2=676$ which is blank as it is an even number.
A: This is a slight variation on the nice answer by @Alex Zorn.
In that answer we find that if the $1$ at the top of the triangle of numbers
is considered entry number $1$ (counting from the left) in row number $1$
(counting from the top), then entry number $k$ in row $n$ is
$n^2−n+2k−1$, and $A$ is entry $k$ in row $n$ where $n,k$ are a solution to
$$3n^2 + n + 6k = 2150.$$
We have $1 \leq k \leq n$ by the way the entries in a row are counted,
so for all $n > 0$ we have
$$3n^2 + n = 2144 \leq 3n^2 + n + 6k = 2150 \leq 3n^2 + 7n = 2150. \tag 1$$
Solving the two equations
$$3n^2 + n = 2144, \tag 2$$
$$3n^2 + 7n = 2150,  \tag 3$$
we see that they each have one positive root.
The positive root of Equation $(2)$ is $r_1 \approx 26.642$ and
the positive root of Equation $(3)$ is $r_2 \approx 25.629$.
But because of Inequality $(1)$, 
whatever the value of $k$ is (provided that $1 \leq k \leq n$)
there is a positive root of $3n^2 + n + 6k = 2150$ 
that is not less than $r_2$ and not greater than $r_1$
(draw a graph if you have trouble seeing this).
The only integer that satisfies both conditions is $26$,
so if $3n^2 + n + 6k = 2150$ has a solution for integer $n$ then
the only possibility is $n = 26$.
Setting $n = 26$, we have $3(26^2) + 26 + 6k = 2150$,
which is a simple linear equation in $k$ with the solution $k = 16$.
We then plug $n$ and $k$ into the formula for $A$ to get the answer.
