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Question

Which of the following has the most number of divisors?


A
99
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B
101
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C
176
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D
182
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Solution

The correct option is D $$176$$
$$99=1\times 3\times 3\times 11$$
$$101=1\times 101$$
$$176=1\times 2\times 2\times 2\times 2\times 11$$
$$182=1\times 2\times 7\times 13$$
So, divisors of $$99$$ are $$1$$, $$3$$, $$9$$, $$11$$, $$33$$, $$99$$
Divisors of $$101$$ are $$1$$ and $$101$$
Divisors of $$176$$ are $$1$$, $$2$$, $$4$$, $$8$$, $$11$$, $$16$$, $$22$$, $$44$$, $$88$$ and $$176$$
Divisors of $$182$$ are $$1$$, $$2$$, $$7$$, $$13$$, $$14$$, $$26$$, $$91$$ and $$182$$.
Hence, $$176$$ has the most number of divisors.

Mathematics

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