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How many ordered pair of integers (*a*, *b*) are needed to guarantee that there are two ordered pairs (*a*_{1}, *b*_{1}) and (*a*_{2}, *b*_{1}) such that *a*_{1} mod 5 = *a*_{2} mod 5 and b_{1} mod 5 = *b*_{2} mod 5? should not answer be 25 here

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Best answer

Based on pigeonhole principle.

You want minimum number of pairs (a,b) to ensure such that

(a_{1},b_{1}) (a_{2},b_{2}) pairs exist in a way

a_{1}=a_{2}mod5 and b_{1}=b_{2}mod5

In a pair(a,b)

a can belong to any of the residue of mod 5(0,1,2,3,4)-Total of 5 values.

Similarly b can take a total of 5 values.

Since a and b can be selected independently, a maximum of total 25(5x5) ordered pairs can be there which can sure that IN NO WAY two pair will exist such that,

(a_{1},b_{1}) (a_{2},b_{2})

a1=a_{2}mod5 and b1=b_{2}mod5

but now since, you have exhausted all the possibilities to make any new pair,now if you make one extra pair, it is guaranteed to have been repeated from the set of 25 ordered pairs formed above.

Hence, you need minimum 26 ordered pairs to ensure that

(a1,b_{1}) (a2,b_{2}) pairs exist in a way

a1=a_{2}mod5 and b1=b_{2}mod5

0 votes

consider this like-there are 5 holes each corresponding to remainders of 5 => 0,1,2,3,4

a1 and a2 will always occupy the same hole

b1 and b2 will always occupy the same hole

a1 and a2 have 5 choices(put them in any one hole). similarly, b1 and b2 have 5 choices(put them in any one hole)

total 5*5=25

$\left \lceil \frac{n}{25} \right \rceil=2$

n=26

a1 and a2 will always occupy the same hole

b1 and b2 will always occupy the same hole

a1 and a2 have 5 choices(put them in any one hole). similarly, b1 and b2 have 5 choices(put them in any one hole)

total 5*5=25

$\left \lceil \frac{n}{25} \right \rceil=2$

n=26