Aptitude Discussion

Q. |
The highest power of 9 dividing 99! completely is |

✖ A. |
12 |

✖ B. |
14 |

✖ C. |
22 |

✔ D. |
24 |

**Solution:**

Option(**D**) is correct

$99! = 99×98×97×96.....×1$

To find the highest power of 9 that divides the above product, it is required to find the sum of powers of all 3's in the expansion

Sum of powers of 3's $=\left[\dfrac{99}{3^1}\right] + \left[\dfrac{99}{3^2}\right] + \left[\dfrac{99}{3^3}\right] + \left[\dfrac{99}{3^4}\right]+...$

where [] is greatest integer.

$=\left[\dfrac{99}{3}\right] + \left[\dfrac{99}{9}\right] + \left[\dfrac{99}{27}\right] + \left[\dfrac{99}{81}\right]+0+...$

$=33+11+3+1$ $= 48$

The highest power of $9 = \dfrac{48}{2} =$ **24.**

**Mohit**

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**Mandar Dharmadhikari**

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The trick is whenever we are asked to find the exact power of a non prime number in such que, we have to go on finding the power using the prime factors of that number and find out which ever power is smaller....

for ex: if 3 and 7 are the prime factors and we ar asked to find out what power of 14 divides x!(say), the we find power for 7 and 3 and consider which ever is smaller.

Thanx....

**Shreya**

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can this question be solved in some other easier way?

Shreya, I guess the question has been solved incorrectly.

Instead of finding the sum of powers of 3, they should have found the sum of powers of 9.

the modified solution then would become:

Sum of powers of 9?s=$[99/9^1]+[99/9^2]+[99/9^3]+ ... $

where [] represents greatest integer.

So the correct answer will become:

$11+1+0+...=12$

So **option A** is the correct choice.

**P.S.** Although there are few mistakes here and there, I still think admin is doing a great job by these questions for free.

wrongly done!...answer should be 14.