Let $2^{a_{1}} + 3^{a_{2}} + 5^{a_{3}} + 7^{a_{4}} + 9^{a_{5}}$ be divisible by 4 where $a_{1}, a_{2}, a_{3}, a_{4}, a_{5}$ are digits from 0 to 9. If largest possible number of values of $(a_{1}, a_{2}, a_{3}, a_{4,} a_{5})$ is $(9000)$ k then k is

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answer is 5.

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Detailed Solution

Let S = $2^{a_{1}} + 3^{a_{2}} + 5^{a_{3}} + 7^{a_{4}} + 9^{a_{5}}$
$= 2^{a_{1}} + {(4 - 1)}^{a_{2}} + {(4 + 1)}^{a_{3}} + {(8 - 1)}^{a_{4}} + {(8 + 1)}^{a_{5}}$
$= 2^{a_{1}} + 4 λ + {(- 1)}^{a_{2}} + {(1)}^{a_{3}} + {(- 1)}^{a_{4}} + {(1)}^{a_{5}} λ \in N$
Case I: When $a_{1} = 0$ in this case
S will either of (4n + 1) or (4n + 3) form which is obviously not divisible by 4.
Case II: $a_{1} = 1, S = 2 + 4 λ + {(- 1)}^{a_{2}} + {(1)}^{a_{3}} + {(- 1)}^{a_{4}} + {(1)}^{a_{5}}$
If S is divisible by 4, last four terms must be equals to 2. For this, one of $a_{2}$ and $a_{4}$ 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: $a_{1} \geq 2, S = 4 n + 4 λ + {(- 1)}^{a_{2}} + {(- 1)}^{a_{3}} + {(- 1)}^{a_{4}} + {(1)}^{a_{5}}$
For S to be divisible by 4, last four terms must be equal to 4 or zero. For this $a_{2}$ and $a_{4}$ must be both even or odd and $a_{3} a_{5}$ 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 $(a_{1}, a_{2}, a_{3}, a_{4}, a_{5})$ is $(9000 \times 5)$
$∴$ k = 5