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What is the minimum number of ordered pairs of non-negative numbers that should be chosen to ensure that there are two pairs (a, b) and (c, d) in the chosen set such that "a ≡ c mod 3" and "b ≡ d mod 5"

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Solution given is like this : 

a = c mod 3 (given) Thus, ‘a’ can be any one of these values : 0, 1, 2 
b = d mod 5 (given) Thus, ‘b’ can be any one of these values : 0, 1, 2, 3, 4 
Thus, ordered pair for (a, b) are : (0, 0), (0, 1), (0, 2), (0, 3), (0, 4), (1, 0), (1, 1), (1, 2), (1, 3), (1, 4), (2, 0), (2, 1), (2, 2), (2, 3), (2, 4) 
Therefore, ordered pair (a, b) has 15 combinations and ordered pair (c, d) has 1 combination. Total combinations = 15 + 1 = 16 
 

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My doubt is cant it be acheived by having just two ordered pairs?? for example:  (2,1) and (8,6) ...so we can have 

a=2,b=1,c=8,d=6 ....and 

"2 ≡ 8 mod 3" and "1 ≡ 6 mod 5"

asked in Set Theory & Algebra by Active (1.1k points) 3 11 24 | 169 views
@Anand . You are choosing the pairs blindly. The, in order to guarantee that both conditions are satisfied, what should be #pairs choosen?
i didn get you ....can u pls elaborate
I mean you need to choose pairs randomly without looking at the pairs.
is this wat the question means???

how many (minimum #)ordered pairs should be there in a set such that if we take any two ordered pairs (a,b) and (c,d) ....then the property holds....

then my set {(2,1),(8,6)} holds the property

Now imagine in this way.

You dont know the conditions to be met.

 

Now, you are supposed to select how many pairs to guarantee that after choosing the pairs(and then you are told the conditions), there will be pairs that meet the conditions?

now i get it .......very well explained ....thanx a lot..... :)

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