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16 Mar 2020, 11:52
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (01:59) correct
43%
(02:09)
wrong
based on 312
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What is the highest power of 5 that will divide the product of first 50 multiples of 5 ?
A. 5^50
B. 5^49
C. 5^62
D. 5^61
E. 5^60
Posted from my mobile device
A. 5^50
B. 5^49
C. 5^62
D. 5^61
E. 5^60
Posted from my mobile device
16 Mar 2020, 12:44
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Raksat
The first 50 multiples of 5 are: 5*1, 5*2, 5*3 ... 5*50
We need to determine the highest power of 5 in: 5*1 x 5*2 x 5*3 x ... x 5*50
= 5^50 * (1 * 2 * 3 * ... * 50)
= 5^50 * 50!
Highest power of 5 in 50! = [50/5] + [50/25] = 10 + 2 = 12
(Note: In this case, the notation above: [x] represents the integer part of the number x, i.e. [2.4] = 2 and [2.9] = 2 as well)
=> Considering the powers of 5, we have: 5^50 * 50! = 5^50 * 5^12 = 5^62
=> Highest power of 5 is 62
Answer C
General Discussion
20 Jul 2021, 05:31
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===========================================================================
Theory: To find the power of a prime number in a factorial we need to divide the factorial (number) by all those powers of the prime number such that those powers are less than the factorial(number)
Example: To find power of 3 in 10! we will proceed as below
\(\frac{10}{3}\) + \(\frac{10}{3^2}\) [we stop at \(3^2\) because \(3^3\) > 10]
= 3 (take the integer just less than the decimal) + 1 = 4
===========================================================================
Coming back to the question
We need to find the power of 5 in first 50 multiples of 5
First 50 multiples of 5 are 5*1, 5*2,...., 5*50
Product of First 50 multiples of 5 = 5*1 * 5*2 * 5*3...* 5*50
Now there are 50 terms, so if we take 5 common in all of them we will get \(5^{50}\)
Product of First 50 multiples of 5 = \(5^{50}\) * (1*2*3*...*50) = \(5^{50}\) * 50!
Now we already know the power of 5 in \(5^{50}\), which is 50, so we just need to find the power of 5 in 50!
Using above theory,
The power of 5 in 50! = \(\frac{50}{5}\) + \(\frac{50}{5^2}\) [we stop at \(5^2\) because \(5^3\) > 50]
= 10 + \(\frac{50}{25}\)
= 10 + 2 = 12
So, total power of 5 in 5*1 * 5*2 * 5*3...* 5*50 = 50 ( for \(5^{50}\)) + 12 (for 50!) = 62
So, answer will be C
Hope it helps!
To learn more about Remainders, watch the following video
Theory: To find the power of a prime number in a factorial we need to divide the factorial (number) by all those powers of the prime number such that those powers are less than the factorial(number)
Example: To find power of 3 in 10! we will proceed as below
\(\frac{10}{3}\) + \(\frac{10}{3^2}\) [we stop at \(3^2\) because \(3^3\) > 10]
= 3 (take the integer just less than the decimal) + 1 = 4
===========================================================================
Coming back to the question
We need to find the power of 5 in first 50 multiples of 5
First 50 multiples of 5 are 5*1, 5*2,...., 5*50
Product of First 50 multiples of 5 = 5*1 * 5*2 * 5*3...* 5*50
Now there are 50 terms, so if we take 5 common in all of them we will get \(5^{50}\)
Product of First 50 multiples of 5 = \(5^{50}\) * (1*2*3*...*50) = \(5^{50}\) * 50!
Now we already know the power of 5 in \(5^{50}\), which is 50, so we just need to find the power of 5 in 50!
Using above theory,
The power of 5 in 50! = \(\frac{50}{5}\) + \(\frac{50}{5^2}\) [we stop at \(5^2\) because \(5^3\) > 50]
= 10 + \(\frac{50}{25}\)
= 10 + 2 = 12
So, total power of 5 in 5*1 * 5*2 * 5*3...* 5*50 = 50 ( for \(5^{50}\)) + 12 (for 50!) = 62
So, answer will be C
Hope it helps!
To learn more about Remainders, watch the following video
