Find The Number Of Trailing Zeros In The Product Of (1^1) GMAT Problem Solving

Question: Find the number of trailing zeros in the product of (1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)

1. 150
2. 200
3. 250
4. 245
5. 225

“Find the number of trailing zeros in the product of (1^1)- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book “GMAT Official Guide”. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. The GMAT Quant topic in the problem-solving part requires calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

It is asked in the question to find out the trailing zeroes in the product of (1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)
It should be noted that there is a trailing zero when we multiply 2 by 5. Now in the given product (1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)
There will be more 5’s than 2’s so the no of two’s will be a limiting factor to find the number of trailing zeros.
We have to count the no of 2’s in even bases.
In (1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)
We have terms 10^10∗20^20∗30^30∗40^40∗50^50
We will only consider the count of two’s in these terms.
10^10∗20^20∗30^30∗40^40∗50^50
=\( 2^{10} *(2^2)^{20} *(2)^{30}*(2^3 )^{40} *(2^{50})\)*terms =\( 2^{10 + 40 + 30 +120+50} \)* terms =
= \(2^{250}\)* terms
Therefore there can be 250 trailing zeros in the given product as there are atleast 250 2’s and more 5’s than 2’s

Correct Answer: C

Approach Solution 2:
It is asked in the question to find out the trailing zeroes in the product of (1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)
Looking at the numbers it looks like
(1x5)^5
(2x5)^10
..
(10x5)^50
There will be more 5’s than the number of 2’s. We know that all the numbers are multiple of 5 but not of 2. Thus, the limiting factor in this case is 2. We are required to find out the number of two in these product terms.
(1^1)∗(5^5)∗(10^10)∗(15^15)∗(20^20)∗(25^25)∗...∗∗...∗(50^50)
Gives ,
2^10 = 10
4^20 = 20 + 20
6^30 = 30
8^40 = 40 + 40 + 40
10^50 = 50
Taking sum of all we get,
120 +50+30+40+10 = 250

Correct Answer: C

Approach Solution 3:

Looking at the numbers it looks like

(1x5)^5
(2x5)^10
...
(10x5)^50

1. Determine the limiting factor. Is it 2 or is it 5? We know that all the numbers are multiple of 5 but not of 2. Thus, the limiting factor in this case is 2. Let's drop all the 5. Then, we count factors of 2 of even multiples.

2^10 = 10
4^20 = 20 + 20
6^30 = 30
8^40 = 40 + 40 + 40
10^50 = 50
=250

Correct Answer: C

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