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Tagged: Answer to the weeks questions

- This topic has 1 reply, 2 voices, and was last updated 1 year, 8 months ago by Mayank Singh.

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- December 3, 2018 at 6:53 am #24575Ashani DasguptaKeymaster
Here are three problems for your brain!

**1**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.)

**2**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.)

**3**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 SinghMemberAns 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

a0+a1+a2…….+a672=0

and( a0+a1+……..+a672)^2=a0^2+a1^2……..+a672^2+2(a0a1+a1a2+a2a3……….a671a672)

0=a0^2+a1^2…….a672^2+2(a0a1………….+a671a672)

a0^2+a1^2………+a672^2=-2(a0a1+………a671a672) (1)

aoa1……..+a671a672<0

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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