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Question

Number of points having position vector ai+bi+ck,a,b,c1,2,3,4,5 such that 2n+3b+5c is divisible by 4 is?


A

140

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B

70

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C

100

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D

75

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Solution

The correct option is B

70


Explanation for the correct option.

Finding the Number of points having position vector ai+bi+ck,a,b,c1,2,3,4,5 such that 2n+3b+5c is divisible by 4 is

Given: position vector ai+bi+ck,a,b,c1,2,3,4,5

Let, 4m=2a+3b+5c

4m=2a+(4-1)b+(4+1)c=2a+4k+(-1)b+(1)c[kisconstant]

If a=1,b=even,c=anynumber

If a1,b=odd,c=anynumber

Therefore, the required number of ways =(1×2×5)+(4×3×5)=70

Hence, option (B) is the correct answer.


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