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Find the digital root of the sum of all positive integers n such that n ! strictly ends in 1 0 0 0 zeroes.
Note- Digital root operation denotes the digital sum operation done over and over again to get a single digit number.
The answer is 1.
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3 solutions
Exactly!!! By the way, nice question once again, @Krishna Ar!!(ah these tags! :\ -_- :?) Your questions are just awesome!!
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No.They aren't because you never fail to solve them. You are an awesome problem solver! and awesome at calculus too @Kartik Sharma
You can solve this to find that n ∈ [ 4 0 0 5 , 4 0 0 9 ] : ⌊ 5 n ⌋ + ⌊ 2 5 n ⌋ + ⌊ 1 2 5 n ⌋ + ⌊ 6 2 5 n ⌋ + ⌊ 3 1 2 5 n ⌋ = 1 0 0 0
yep, just what i did!
You can use the equation to find the numbers ending with 1000 zeroes.
⌊ 5 n ⌋ + ⌊ 2 5 n ⌋ + ⌊ 1 2 5 n ⌋ + ⌊ 6 2 5 n ⌋ + ⌊ 3 1 2 5 n ⌋ = 1 0 0 0 .
You will get an approx answer = 4001.
But we are short of 1 zero Hence the number should be 4005.
And the other values would be 4005, 4006, 4007, 4008, 4009.
S u m = 2 0 0 3 5
So digital root = 1 .
We only need to consider 5s in these questions as we have plenty of 2s.
When we find 4000!, it has 800 fives, 160 25s, 32 125s, 6 625s, and 1 3125. Add them to get 999
therefore. so 4005!,4006!....4009! has 5^1000, trivially 10^1000 as factor.
20035, 10, 1.
UNITY
Factorial of numbers from 4005 to 4009 have strictly 1000 zeros as they have 5^1000 as their factor. Digital sum of these is 1. ANSWER = 1.