# Please solve this riddle in easy way

Riddle: there is a four story building let's consider some number of people comes and settle in all 4 floors .we don't know how many people comes and also don't know how many people are settle in all 4 floors (means in 1,2,3 and 4) . But we have a clue that 4th floor have crowded then to reduce the people in fourth floor the peoples in all other floors get doubled and from 4th floor same number of people removed means people in 1st+people in 2nd and people in 3rd after all this now 3rd floor is crowded and same done with this like 4th after this 2nd floor crowded and same done with 2nd as 4th and after this 1st floor is crowded but after doing same with 1st floor all floor have same number of people now solve how many people was initially in 1,2,3 and 4 floor and tell total people i know answer please do this MATHEMATICALY

## 1 Answer

First notice that the solution is not unique ; if we have any solution, then replacing multiplying the numbers with any other number we get again a solution. (One can just replace "person" by "couple" and have twice as many persons everywhere.) So we are looking for the smallest possible solution.

Let's say in the end we have E people in each floor, so total number = 4E. The last step was that people in 2nd 3rd 4th floor were doubled and this number was removed from 1st floor, so the other floors had E/2 people and 1st floor had E + E*3/2 = E*5/2.

This was the result when the same was done with the 2nd floor, so before that the 3rd and 4th floors had E/4 and 1st had E*5/4 and the 2nd floor had E/2 + E*(5 + 1 + 1)/4 = E*9/4 people.

This was the result after the same was done for the 3rd floor, so before the 4th floor had E/8 and the 1st had E*5/8 and the 2nd had E*9/8, and all this was removed from the 3rd floor which therefore had E/4 + E*(5 + 9 + 1)/8 = E*17/8.

This was the result after the same was done for the 4rd floor, so the 1st floor had E*5/16 and the 2nd had E*9/16 and 3rd had E*17/16, and all this was removed from the 4th floor which therefore had E/8 + E*(5+9+17)/16 = E*33/16 people, initially.

All of this must be integers, so we must have E a multiple of 16 ; least possibility is E=16 on each floor in the end, for a total of 64 people in the buildin.

That means that initially we had 5, 9, 17 and 33 people on the 1st, 2nd, 3rd & 4th floor (again total = 14+50 = 64), and after the first step we would have (10, 18, 34) people on the 1st, 2nd & 3rd floor and 5+9+17 = 31 less on the 4th floor, which means only 2 on the 4th floor.

Then after the second step we have (20, 36, 4) on the 1st, 2nd and 4th floor and 10+18+2 = 30 less than 34, i.e., also 4 on the 3rd floor.

And after the 3rd step we have (40, 8, 8) on the 1st, 3rd and 4th floor and 20+4+4 = 28 less than 36, i.e., also 8 on the 2nd floor.

And finally, after the last step, we have 16 on each of the 2nd, 3rd and 4th floor and 3*8 = 24 less than 40, i.e., also 16 on the 1st floor, as expecte

- 1 Answer
- 259 views
- Pro Bono

### Related Questions

- Prove that $V={(𝑥_1,𝑥_2,⋯,𝑥_n) \in ℝ^n ∣ 𝑥_1+𝑥_2+...+𝑥_{𝑛−1}−2𝑥_𝑛=0}\}$ is a subspace of $\R^n$.
- Find the null space of the matrix $\begin{pmatrix} 1 & 2 & -1 \\ 3 & -3 & 1 \end{pmatrix}$
- Stuck on this and need the answer for this problem at 6. Thanks
- Find $x$ so that $\begin{pmatrix} 1 & 0 & c \\ 0 & a & -b \\ -\frac{1}{a} & x & x^2 \end{pmatrix}$ is invertible
- Linear Algebra - Vectors and Matrices
- Certain isometry overfinite ring is product of isometries over each local factor
- Closest Points on Two Lines: How to use algebra on equations to isolate unknowns?
- Let $\mathbb{C} ^{2} $ a complex vector space over $\mathbb{C} $ . Find a complex subspace unidimensional $M$ $\subset \mathbb{C} ^{2} $ such that $\mathbb{C} ^{2} \cap M =\left \{ 0 \right \} $