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The minimum number of non negative integers we have to choose randomly so that there will be atleast two integers x and y such that x+y or x-y is divisible by 10

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If a number is divided by 10 then it leaves the remainders { 0,1,2,3,...,9}

Observe that for two nos. x and y of the forms 10k + p and 10k + q respectively. (Here p and q both are less than 10)

1) (x+y) is divisible by 10 iff (p+q) = 10

2) (x-y) is divisible by 10 iff (p-q) = 0 i.e. p=q.


Now, suppose we take 6 nos. of the form {10k, 10k+1, 10k+2, 10k+3, 10k+4, 10k+5}

Observe that in the above nos. neither p=q nor (p+q) = 10

So, we will not get two integers x and y from the above set such that either (x+y) or (x-y) is divisible by 10.

So, the minimum number of nonnegative integers required to satisfy the given condition must be greater than 6.

But if add an extra no. of the form 10k + r to the above 6 numbers then we will have two cases

Case 1: r belongs to the set {0,1,2,3,4,5}

In this case we will get two numbers x and y such that p=q.

So, (x-y) is divisible by 10.

Case 2: r does not belong to the set {0,1,2,3,4,5}

In this case we will get two numbers x and y such that (p+q) = 10

So, (x+y) is divisible by 10.

So, if we take arbitrary 7 nonnegative integers then we must get at least two numbers x and y such that either (x+y) or (x-y) is divisible by 10.

Hence,the required answer is 7.

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