Tagged: Answer to the weeks questions
- December 3, 2018 at 6:53 am #24575Ashani DasguptaKeymaster
Here are three problems for your brain!
Can you glue edges of an octagon to make a two-holed torus!
(This week we are working on this idea in a geometry session at Cheenta.)
We want to take a 672 degree polynomial with integer coefficients. Suppose we plugin 673th root of 1 in that polynomial and get 0.
What can you say about the integer coefficients of this polynomial?
(This problem will come up in a Complex Number and Geometry session this week.)
We want to partition the number of shortest paths from (0,0) to (2019, 2019). You are allowed to walk only on the grid (lines with either x coordinate an integer or y coordinate an integer between 0 to 2019)
Can you take the vertical line x = 673 to partition these set of shortest paths into mutually exclusive and exhaustive subsets of paths?
(This problem will come up in a Combinatorics session this week.)December 3, 2018 at 3:22 pm #24615Mayank SinghMember
Ans Q2. The answer to problem 2 will be that all the coefficients are 0 because let the polynomial be
a0+a1x+a2x^2……….a672x^672 . If we plug 1
only case possible is a0=-a1, a2=-a3………..a670=-a671 wher a1,a3…..a671 are ve integers. so we get a672 as a remainder hence the sum is not equal to 0.
SO the only case possible is that all the coefficients are 0.
If a0=-(a1+a2) for the least and a3=-(a4+a5) and so on then in (1) RHS will be negative and LHS positive which is not possible.
Ans Q1. yes
First join the edges of octagon to form a rectangle. Then make a join the four vertices to opp sides of the rectangle to form a 3d pipe like figure. Then extend the edges of the pipe to provide a space for inserting the pipe like structure formed. So when we extend the pipe like structure and insert it in the space a two holed torus will be formed.
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