Select Page

# Problems for the week!

Home Forums Math Olympiad, I.S.I., C.M.I. Entrance Problems for the week!

Viewing 2 posts - 1 through 2 (of 2 total)
• Author
Posts
• #24575
Ashani Dasgupta
Keymaster

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.)

#24615

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

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.

Viewing 2 posts - 1 through 2 (of 2 total)
• You must be logged in to reply to this topic.