What is the largest number of positive integer to be picked up randomly so that the sum or difference of any two of the chosen number is divisible by 10?
Step-by-step Solution:
The question is a classic application of the Pigeonhole Principle and is best understood as: "What is the minimum number of integers one must select to guarantee that for at least one pair of the chosen numbers (let's call them 'a' and 'b'), their sum (a+b) or their difference (a-b) is divisible by 10?"
The divisibility of the sum or difference depends on the remainders of the numbers when divided by 10. If the remainders are 'r_a' and 'r_b', the condition is met if either $r_a = r_b$ or $r_a + r_b = 10$.
We can sort the 10 possible remainders (0 to 9) into groups or "pigeonholes". If any two numbers have remainders from the same group, the condition is satisfied.
There are 6 such groups:
To avoid finding a pair that satisfies the condition, one could pick one number from each distinct group. This allows for a maximum of 6 picks without a guaranteed match (the "worst-case scenario").
However, when you pick the 7th number, its remainder must belong to one of the 6 groups you have already picked from. This guarantees that you will have two numbers from the same group, which in turn guarantees that their sum or difference will be divisible by 10.
Thus, the required number is 7.