# A math problem is able to have multiple right solutions

What would be your answer to the question mark? The most common answers are 40 and 96... Does anyone could say why? Hence if it is given the first terms in a sequence and the sequence is not well defined, the next term is able has multiple right solutions, even infinite solutions. For instance, in the next problem Find the solution which I gave one solution, is able to have infinite answers, even the other answers given in the problem can be right... Just only have to take polynomials with two variables x,y which satisfies the conditions, for example,can you find a polynomial with variables x, y satisfying the first conditions and such that f(13,4) = 123 or f(10,9) = 123?... So my conclusion is this one: Can you post one riddle where different right solutions are valid?and why my "riddle" is able to have 40 or 96 as valid solutions?can you find other right solution and say why?

4 years, 5 months ago

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I think they expect 96 as the answer if they do this. $a+b=a×(b+1)$ But like @Nihar Mahajan said you can have $\infty$ solutions

- 4 years, 5 months ago

The riddle has infinitely many solutions using LZOB or Lagrange interpolation, since the choices are not provided.

- 4 years, 5 months ago

Why are you assuming the function is a polynomial? (just asking)

- 4 years, 5 months ago

Its not wrong to assume a polynomial right?

- 4 years, 5 months ago

I could be any arbitrary function of two variables for that matter as we've not been given that the function IS a polynomial.

- 4 years, 5 months ago

So it can be $\infty$ if its a polynomial and if its some other function it can have finite or infinite. Therefore in general it can be said infinite

- 4 years, 5 months ago

- 4 years, 5 months ago

Hey I was asking @Deeparaj Bhat. And it was a comment and not a question

- 4 years, 5 months ago

ok, sorry, my apologies

- 4 years, 5 months ago

- 4 years, 5 months ago

How is the answer 96 or 40?

- 4 years, 5 months ago

I got 96 but not 40.

- 4 years, 5 months ago

ok, 96 can be got of this form, $\begin{cases}1 + 4 = 1 \cdot 4 + 1 = 5 \\ 2 + 5 = 2 \cdot 5 + 2 = 12 \\ 3 + 6 = 3 \cdot 6 + 3 = 21\end{cases}$ so I can say $8 + 11 = 8 \cdot 11 + 8 = 96$. Now, we are going to get 40,ok...please, be cal, wait my answer,haha..

- 4 years, 5 months ago

Wait, I will wait. (LOL)

- 4 years, 5 months ago

Now, we are going 40.. We can consider (1,4), (2,5), (3,6) and (8,11) to be the first terms of one sequence in $\mathbb{N}^2$ and $1 + 4 = 5$ and we can define for the next terms its sum added to the sum obtained before. Hence $2 + 5 = 7 + \text{ sum obtained before } = 7 + 5 = 12$ , now $3 + 6 = 9 + \text{ sum obtained before } = 9 + 12 = 21$ so $8 + 11 = 19 + \text{ sum obtained before } = 19 + 21 = 40$

- 4 years, 5 months ago

Oh! I considered $8,11$ as the eighth term!

- 4 years, 5 months ago

but, they can be considered the first terms in a sequence. Now I'm going with the Lagrange interpolation

- 4 years, 5 months ago

- 4 years, 5 months ago

- 4 years, 5 months ago

We have $f(5,3) = 28, \space f(12,10) = 222 \space , f(9,1) = 810, \space f(4,2) = 26, \space f(8,3) = 511, f(10,1) = 911, \space f(7,3) = 410$. This problem using Lagrange interpolation is able to have $\infty$ answers. For example, I want furthemore $f(13,9) = 123.$ Is evrything allright so far? haha. Ok, I'm going to look for a polynomial with variables x,y and some constants a,b,c,d,e,f,g,h,i, ...fullfiling these conditions,(notice I'm going to repeat this factor (y -1) and (y - 3), I would not to do it, but I'm going to do it for clarity) so the polynomial has to be $\small f(x,y) = a(x -5)(y -3)(x -12)(y -10)(x - 9)(y -1)(x - 4)(y -2)(x - 8)(y -3)(x - 10)(y -1)(x - 7)(y - 3) +$ $\small + b(x -5)(y -3)(x -12)(y -10)(x - 9)(y -1)(x - 4)(y -2)(x - 8)(y -3)(x - 10)(y -1)(x - 13)(y -9) +$ $\small + c(x -5)(y -3)(x -12)(y -10)(x - 9)(y -1)(x - 4)(y -2)(x - 8)(y -3)(x - 7)(y - 3)(x -13)(y - 9) +$ $\small + d(x -5)(y -3)(x -12)(y -10)(x - 9)(y -1)(x - 4)(y -2)(x - 10)(y -1)(x - 7)(y - 3)(x - 13)(y - 9) +...$ and get the constants a,b ,c ,d e,f, g, h,... fulfilling the requisites... so this problem has $\infty$ right solutions... Ok, now I'm going to eat... Later I'll review the comments

- 4 years, 5 months ago

Exactly, for the problem Find the solution you can use Lagrange interpolation with two variables, please wait a bit, and I'll give the answer

- 4 years, 5 months ago

52

Assuming there is something fundamental operation that isn't explicit, just like the assumption that the last answer is also added to the next equation... Goes like this... The pattern is multiplying the second number secretly by a sequential while number, just like secretly having the last answer added to the next equation...

1+(4[×1])=5 2+(5[×2])=12 3+(6[×3])=21 8+(11[×4])=52

This makes all the answers correct.

- 3 years ago

## 8+(11[×4])=5

- 3 years ago

In this riddle the answer of the previous statement is added to the terms of the original statement.
First statement starts with 1 + 4 = $\color{#D61F06}{\text{5}}$.
Second statement:- $\color{#D61F06}{\text{5}}$ + 2 + 5 = $\color{#3D99F6}{\text {12}}$.
Third statement:- $\color{#3D99F6}{\text {12}}$ + 3 + 6 = $\color{#20A900}{\text {21}}$.
Fourth statement should be $\color{#20A900}{\text {21}}$ + 8 + 11 = $\boxed{40}$.

- 4 years, 5 months ago

## 3+(6[×3])=21

8+(11[×4])=52

- 3 years ago

That was supposed to be a clarification of my own post... Sorry. 'Doing this from my phone while walking to the store.

- 3 years ago