The number of distinct primes dividing $12! + 13! + 14!$ is

5

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6

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7

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8

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Solution

The correct option is D $5$
$12! + 13! + 14!$
$12! (1 + 13 + 14 \times 13)$
$12! \times 196$
Which is only divided by possible distinct primes
$2, 3, 5, 7, 11$

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divides

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Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

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(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

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The number of distinct primes dividing 12!+13!+14! is

The correct option is D 512!+13!+14!12!(1+13+14×13)12!×196Which is only divided by possible distinct primes2,3,5,7,11

The number of distinct primes dividing $12! + 13! + 14!$ is

The correct option is D $5$
$12! + 13! + 14!$
$12! (1 + 13 + 14 \times 13)$
$12! \times 196$
Which is only divided by possible distinct primes
$2, 3, 5, 7, 11$