What is the Largest Power of 3 Contained in 200! GMAT Problem Solving

Question: What is the largest power of 3 contained in 200!

1. 88
2. 48
3. 66
4. 97
5. 39

“What is the largest power of 3 contained in 200!” - is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. 
In the GMAT Problem Solving section, examiners measure how well the candidates make analytical and logical approaches to solve numerical problems. In this section, candidates have to evaluate and interpret data from a given graphical representation. In this section, mostly one finds out mathematical questions. Five answer choices are given for each GMAT Problem solving question. 

Solution and Explanation: 

Approach Solution 1:

 \(\frac{200}{3}+\frac{200}{3^2}+\frac{200}{3^3}+\frac{200}{3^4} \)

\(=66+22+7+2=97\)

Considering the given case to find the largest power of 3 in 200, the question can be solved first by dividing 200 by 3.
This can be then added with 200 divided by squares of 3.
Further, it is added with 200 by a cube of 3.
Finally, add 200 divided by 3 to the power of 4.

The values derived from each of them include 66, 22, 7, and 2. Adding all the values of 66, 22, 7, and 2, the result is 97 hence, D is the right answer.

Correct Answer: D

Approach Solution 2:

\(\Rightarrow \frac{200}{3}=66 \)

\(\Rightarrow \frac{66}{3}=22\)

\(\Rightarrow \frac {22}{3} = 7\)

\(\Rightarrow \frac {7}{3}=2\)

To find the answer to the largest power of 3 contained in 200, the first equation can be identified with 200 by 3 which equals 66.
This further implies for citing the value 66 to be divided 3 equal to 22.
This implies that 22 can be divided by 3 to get the answer 7.
Finally, when 7 is divided by 3 the value appears as 2.
By adding all the values found through these formulas, the resultant answer is 97 which is option D.

Correct Answer: D

Approach Solution 3:

Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on

Here [n/x] is greatest Integer value of (n/x) less than or equal to (n/x)
Example: [100/3] = [33.33] = 33

Where,
[n/x] = No. of Integers that are multiple of x from 1 to n
[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in the previous step and second is being counted at this step
[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in the previous two-step and whose third power is counted at this step
And so on.....

Power of prime 3 in 200! = [200/3] + [200/3^2] + [200/3^3] + [200/3^4] + ... and so on

Power of prime 3 in 100! = 66+ 22 + 7 + 2 + 0... and so on = 97

Correct Answer: D

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