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Division of a Polynomial by a Binomial

Consider the ...

Question

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

1

only

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2

only

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Both

1

and

2

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Neither

1

nor

2

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Solution

The correct option is A $1$ only
$a_{1} a_{2} a_{3} a_{4} a_{5} = 10000 a_{1} + 1000 a_{2} + 100 a_{3} + 10 a_{4} + a_{5}$
$\Rightarrow (9999 a_{1} + 999 a_{2} + 99 a_{3} + 9 a_{1}) + (a_{1} + a_{2} + a_{3} + a_{4} + a_{5})$
$∴$ for $a_{1} a_{2} a_{3} a_{4} a_{5}$ to be divisible by 'g'
$(a_{1} + a_{2} + a_{3} + a_{4} + a_{5})$ should be divisible by $9$

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Similar questions

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

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}

x + 1

f (x) = a_{0} x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + \dots + a_{n} = 0

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