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number of ways of selecting two numbers from 1 to 12 such that their sum is divisible by 3.
Dear Naveen for a no to be divisible by 3 , possibilities are 3,6,9,15,18,21 if 2 no. are x and y , den eqns are x + y = 3 ,x + y = 6, x + y = 9, x + y = 12 , x + y = 15 , x + y = 18 , x + y = 21 for 1st eqn x,y wud be 1 & 2 for 2nd set is (2,4),(1,5) for 3rd (1,8),(2,7),(3,6),(4,5) for 4th (1,11),(2,10),(3,9),(4,8),(5,7),(6,6) for 5th (12,3),(11,4),(9,6),(10,5),(8,7) for 6th (12,6),(11,7),(10,8) for 7th (10,11) just count the possibility and dat is ur ans
Dear Naveen
for a no to be divisible by 3 , possibilities are 3,6,9,15,18,21
if 2 no. are x and y , den
eqns are
x + y = 3 ,x + y = 6, x + y = 9, x + y = 12 , x + y = 15 , x + y = 18 , x + y = 21
for 1st eqn x,y wud be 1 & 2
for 2nd set is (2,4),(1,5)
for 3rd (1,8),(2,7),(3,6),(4,5)
for 4th (1,11),(2,10),(3,9),(4,8),(5,7),(6,6)
for 5th (12,3),(11,4),(9,6),(10,5),(8,7)
for 6th (12,6),(11,7),(10,8)
for 7th (10,11)
just count the possibility and dat is ur ans
12
48
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