If N is the product of even numbers from 2 to 100, and N is divisible by 4725^n , find the largest possible value of n
Answers
Given : N is the product of even numbers from 2 to 100, and N is divisible by 4725^n
To find : largest possible value of n
Solution:
4725 = 3³ * 5² * 7
4725ⁿ = 3³ⁿ * 5²ⁿ * 7ⁿ
even numbers having 7 as factor
14 , 28 , 42 , 56 , 70 , 84 , 98
98 has 7 * 7 * 2
so total 8 times multiplier of 7
max n = 8
even number having factor 5
10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , 90 , 100
50 has 5 * 5 * 2
100 has 5 * 5 * 4
so total 12 times multiplier of 5
Hence 2n = 12
=> max n = 6
even numbers having 3 as factor
6 , 12 , 18 ................................................., 96
18 = 3 * 3 * 4
36 = 3 * 3 * 4 and so on
Much greater than 18
Hence n > 6
n = 6 is the largest possible value of n
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