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Divisibility by 3

A number a1...

Question

A number $a_{1} a_{2} a_{3} a_{4} a_{5}$ is divisible by 9 if
(i) $a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$ is divisible by $9.$
(ii) $a_{1} - a_{2} + a_{3} - a_{4} + a_{5}$ is divisible by $9.$
Which of the above statements is/are correct?

only (i)

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only (ii)

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both (i) and (ii)

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neither (i) nor (ii)

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Solution

The correct option is A only (i)
The given number is divisible by $9$ , if sum of
its digits is divisible by $9$ .
$∴ a_{1} + a_{2} + a_{3} + a_{4} + a_{5}$ is divisible by 9
So, option A is correct.

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Similar questions

a_{1} a_{2} a_{3} a_{4} a_{5}

9

a_{1} + a_{2} + a_{3} + a_{4} + a_{5}

9

a_{1} - a_{2} + a_{3} - a_{4} + a_{5}

9

a_{1} a_{2} a_{3} a_{4} a_{5} a_{6}

(a_{1} + a_{3} + a_{5}) - (a_{2} + a_{4} + a_{6}) = 0

(a_{2} + a_{4} + a_{6}) - (a_{1} + a_{3} + a_{5}) = 0

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}

a_{2}

a_{3}

a_{2}

a_{3}

a_{4}

a_{3}

a_{4}

a_{5}

a_{1}

a_{3}

a_{5}

a_{1}, a_{2}, a_{3}, a_{4}, a_{5}

6 x^{5} - 41 x^{4} + 97 x^{3} - 97 x^{2} + 41 x - 6 = 0,

| a_{1} | \leq | a_{2} | \leq | a_{3} | \leq | a_{4} | \leq | a_{5} |,

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