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Q.

Let 2a1+3a2+5a3+7a4+9a5 be divisible by 4 where a1,a2,a3,a4,a5 are digits from 0 to 9. If largest possible number of values of (a1,a2,a3,a4,a5) is (9000) k then k is 

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answer is 5.

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Detailed Solution

Let S = 2a1+3a2+5a3+7a4+9a5  
=2a1+(41)a2+(4+1)a3+(81)a4+(8+1)a5 
=2a1+4λ+(1)a2+(1)a3+(1)a4+(1)a5λN 
Case I: When a1=0 in this case 
S will either of (4n + 1) or (4n + 3) form which is obviously not divisible by 4.
Case II: a1=1,  S=2+4λ+(1)a2+(1)a3+(1)a4+(1)a5 
If S is divisible by 4, last four terms must be equals to 2. For this, one of a2 and a4 must be odd and other to be even. This can be done in 2(1 x 5 x 10 x 5 x 10) ways = 5000 ways.
Case III: a12, S=4n+4λ+(1)a2+(1)a3+(1)a4+(1)a5  
For S to be divisible by 4, last four terms must be equal to 4 or zero. For this a2 and a4 must be both even or odd and a3  a5 can be any digit. This can be done in 
2 x ( 8 x 5 x 10 x 5 x 10) = 40,000 ways 
 largest possible number of values of (a1,a2,a3,a4,a5) is (9000×5)  
 k = 5 

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