How many ordered pairs of integers (a
b.Are needed to guarantee that there are two ordered pairs??
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hey bro here is your answer. .
For each a,ba,b in (a,b)(a,b), there are 5 different congruence classes in mod5mod5. To answer the question of How many ordered pairs of integers of the form (a,b)(a,b)?, we need to consider 25 different congruence classes that both a,ba,b will form. Thus the problem is now reduced to pigeon hole problem,
⌈n25⌉=2⌈n25⌉=2
By solving the inequality of of the ceil function (I don't have to but just to be formal),
⌈n25⌉−1<n25≤n25⌈n25⌉−1<n25≤n25
2−1<n25≤22−1<n25≤2
25<n≤5025<n≤50
Thus have to be 25+1=2625+1=26 by the Pigeon hole principle.
please marke brainlist.
For each a,ba,b in (a,b)(a,b), there are 5 different congruence classes in mod5mod5. To answer the question of How many ordered pairs of integers of the form (a,b)(a,b)?, we need to consider 25 different congruence classes that both a,ba,b will form. Thus the problem is now reduced to pigeon hole problem,
⌈n25⌉=2⌈n25⌉=2
By solving the inequality of of the ceil function (I don't have to but just to be formal),
⌈n25⌉−1<n25≤n25⌈n25⌉−1<n25≤n25
2−1<n25≤22−1<n25≤2
25<n≤5025<n≤50
Thus have to be 25+1=2625+1=26 by the Pigeon hole principle.
please marke brainlist.
vasanthij97:
r u ajay!?
Answered by
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- ⌈n25⌉=2
- By solving the inequality of the ceil function,
- ⌈n25⌉−1<n25≤n25
- 2−1<n25≤2
- 25<n≤50
- 25+1=26 by Pigeon hole principle.
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