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"