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In the following form of odd numbers enter image description here

If the numbers enter image description here

taken from the form where $A+B+C=2149$

Find $A$

any help will be appreciate it, thanks.

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  • $\begingroup$ Hint: looking at the numbers in the picture, try to find a pattern that tells you what $B$ and $C$ are in terms of $A$. Then plug $B$ and $C$ into $A+B+C=2149$ and solve for $A$. It helps to give each line a number, so the line with 1 is line number 1, the line with 3 and 5 is line number 2, the line with 7, 9, and 11 is line number 3, etc. $\endgroup$ – wltrup Aug 4 '15 at 22:50
  • $\begingroup$ What interesting puzzles you come up with! $\endgroup$ – hypergeometric Aug 6 '15 at 9:33
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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, $T(25)=325<\frac{681+1}{2}=341<351=T(26)$

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  • $\begingroup$ Oh man, you're genies, that's correct, because there is 4 options, and 681 one of them, thanks $\endgroup$ – Oiue Aug 4 '15 at 23:08
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    $\begingroup$ +1 for the ballpark paragraph. I stress this when I tutor students, that having a ballpark estimate (even one it takes $5$ seconds to come up with) and compare that to the exact answer you got at the end is a very effective way to see if you've done any big mistakes. $\endgroup$ – Arthur Aug 4 '15 at 23:16
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$A$ 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 $A = 26^2 - 26 + 32 - 1 = 681$.

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  • $\begingroup$ I finished this off a little differently (my solution is a mere variation of yours, as I mention there), but I liked your derivation of the equation $3n^2 + n + 6k = 2150$, which is the key element of either solution. I would have added my variation as a comment on your answer but it didn't look like it would fit the allotted space. $\endgroup$ – David K Aug 5 '15 at 5:30
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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!

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  • $\begingroup$ If you "fill in" the table so that, for example, row $4$ is 13 14 15 16 17 18 19, then the middle A is still $x^2$ and $A+B+C = 3x^2+4x+2$. Solving $3x^2+4x+2=2149$ for $x$, we get $x=26.094$. So we are in row $26$ and our A is just to the right of $A=26^2$, where $A+B+C=2134$. In the filled in table, each shift to the right adds $1$ to $A, B,$ and $C$ and adds $3$ to $A+B+C$. $(2149-2134)/3=5$. So $A = 26^2 + 5 =681$ $\endgroup$ – steven gregory Aug 5 '15 at 14:56
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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 $B=2*(341+25+1)-1=733$, $C=735$.

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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}

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  • $\begingroup$ Ok, you just should edit the first row $a=1, b=3, c=5, a+b+c=9$ $\endgroup$ – Oiue Aug 5 '15 at 0:08
  • $\begingroup$ @Oiue. Yep. It's nine in my notes too. Thanks. $\endgroup$ – steven gregory Aug 5 '15 at 1:18
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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.

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  • $\begingroup$ every solutions her always adds something new, thanks :) $\endgroup$ – Oiue Aug 5 '15 at 18:41
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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.

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