practice problems pending
Q. An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is 34 . What is the value of the cube?
Solution:
Each vertex is part of 3 edges.
Each edge is part of 2 faces.
So each vertex gets added 6 times.
34*6 = 204 = Answer
Q. Suppose you are asked to fill up a 5×5 grid with any 25 natural numbers such that the sum of numbers in each row is even and each column is odd. Do there exist such a configuration?
Solution:
No.
If each row sum is odd, then total grid sum = odd + odd + odd + odd + odd = 5 odds added = odd
But by the same logic for columns it's even.
So no.
Q.
Solution:
Factorize:
How?
Try with (x^(2^3) - y^(2^3))
which is
x^8 - y^8
= (x^4 - y^4) (x^4 + y^4)
= (x^(2^2) - y^(2^2))(x^(2^2) + y^(2^2))
So,
(x^(2^2024) + y^(2^2024)) =
(x-y)(x+y)(x^2+y^2)(x^4+y^4)...(x^(2^2023) + y^(2^2023))
Now we will count number of 2's in each term.
x-y = 8k + 2 = 2(4k+1) = 2*odd => one 2
x +y = (y + 8k + 2) + y = 2(y + 4k + 1)
y is odd = 2m + 1
=> x + y = 2(2m + 1 + 4k + 1) = 4(m + 2k + 1) => two 2's
Next:
x^2 + y^2
Useful property here: square of an odd number is always of the form 8k + 1.
Why?
(2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m+1) + 1. Here m.(m+1) is always even. So it's 8k + 1
So x^2 + y^2 = 8p + 1 + 8q + 1 = 2(4p + 4q + 1) => one 2.
Similarly all the next terms will yield on 2 since each of them is sum of 2 odd numbers squared.
There are 2023 such terms.
So total 2's = 1 + 2 + 2023 = 2026 = answer.
Q. Given a,b,c are integers and ab + bc + ca = abc - a - b - c
is it necessary that each of a,b,c will be even?
Solution:
(1+a)(1+b)(1+c) = 1 + a + b + c + ab + bc + ca + abc
We are given:
a + b + c + ab + bc + ca = abc
=> 1 + a + b + c + ab + bc + ca + abc= 1 + 2abc
=> (1+a)(1+b)(1+c) = 1 + 2abc
RHS = odd
So LHS has to be odd as well.
Even if one of a,b,c is odd LHS will become even.
So each of them has to be even.
Q.
Solution:
Let's check by 3 first.
Case 1:
At least one of a,b,c is a multiple of 3.
Then the entire expression is divisible by 3.
Case 2:
None of a,b,c is divisible by 3.
Then by pigenhole principle at least 2 of them share the same remainder by 3 (1,2).
Also we know that n^3 = n mod 3 for every integer.
Since n^3 - n = n(n-1)(n+1) and product of 3 consecutive integers is always divisible by 3.
WLOG assume that a,b share same remainder => a^3 = b^3 mod 3 => a^3 - b^3 = 0 mod 3
So the entire expression is again divisible by 3.
So both the cases are covered. And 3 is the answer.
Let's test by 5 as well.
For a=1,b=2,c=3 it's not divisible by 5. Counter example.
Let's test by 7 as well.
Case 1:
One of a,b,c is divisible by 7.
Done.
Case 2:
None of them is divisible by 7.
Now consider cubes mod 7.
1^3 mod 7 = 1
2^3 mod 7 = 8 mod 7 = 1
4^3 mod 7 = 16*4 mod 7 = 8 mod 7 = 1
6^3 mod 7 = 1 * 6 mod 7 = 6
So by pigenhole principle again, 2 of the three cubes will share the same remainder.
And the entire expression is divisible by 7.
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