Can you show that given any 52 integers there exists 2 of them whose sum or otherwise difference is divisible by 100?

since all remainders r should be different and r<100, we can pick 2 safe remainders: 0 and 50 – that is 2 values among the 52 can safely have these remainders. there are 49 remainder pairs which produce sums of 100: 1+99, 2+98., 49+51. So, there are.

How many integers that one must choose to ensure that among the chosen ones there are at least two numbers whose difference is divisible by 2021?

Here m is an odd number because all the 2’s in the factors of n is in 2k. Hence,there are 100 values for m. m is equal in both the numbers. So, among the chosen 101 integers there must exist at least two integers such that one is divisible by the other.

How to find the number of values in a given range divisible by a given value?

14 Answers

1. find the first value after x that is divisible by z . You can discard x : x_mod = x \% z; if(x_mod != 0) x += (z – x_mod);
2. find the last value before y that is divisible by y . You can discard y : y -= y \% z;
3. find the size of this range: if(x > y) return 0; else return (y – x) / z + 1;

What is the number of students required in DSGT class to be sure that at least six will receive the same grade if there are five possible grades A B C D E?

26
Thus, 26 is the minimum number of students needed to ensure that at least six students will receive the same grade.

How do you prove pigeonhole principle?

(The Pigeonhole Principle, simple version.) If k+1 or more pigeons are distributed among k pigeonholes, then at least one pigeonhole contains two or more pigeons. Proof. The contrapositive of the statement is: If each pigeonhole contains at most one pigeon, then there are at most k pigeons.

How many students in a class must there be to ensure that 6 students get the same grade?

Plugging k and r into this equation gives us N = 5(6‐1)+1 = 26. Thus, 26 is the minimum number of students needed to ensure that at least six students will receive the same grade.

How do you write a pigeonhole proof?

Pigeonhole Principle: If k is a positive integer and k + 1 objects are placed into k boxes, then at least one box contains two or more objects. Proof: We use a proof by contraposition. Suppose none of the k boxes has more than one object. Then the total number of objects would be at most k.