Question 1023806: What is the least positive integer n such that n! is divisible by 1000?
The sum of the first five terms of an arithmetic sequence is 90 less than the sum of the next five terms. What is the absolute difference between two consecutive terms of this sequence? Express your answer as a common fraction.
Found 3 solutions by Edwin McCravy, msalonga, AnlytcPhil:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website!
1000 = 5*5*5*2*2*2
To have a factorial divisible by 1000, it must have 3 factors of 5
and 3 factors of 2.
Every other integer has a factor of 2 so there are plenty of those,
so we must make sure that we have 3 factors of 5 somewhere in the
products. Let's start buiding our factorial
1*2*3*4*5 <-- that only has 1 factor of 5
(It already has 3 factors of 2)
So we keep putting on more factors:
1*2*3*4*5*6*7*8*9*10 <-- That only has 2 factors of 5
So we keep going once more:
1*2*3*4*5*6*7*8*9*10*11*12*13*14*15 <-- that's it!
That's 15!. It is the smallest factorial that is divisible by 1000.
Notice that 14! = 87178291200 won't do.
It only has two 0's on the end, so it's only divisible
by 100. But
15! = 1307674368000
is the smallest factorial that has 3 0's on the end, which
a number must have to be divisible by 1000.
Answer: 15! = 1307674368000
Edwin
Answer by msalonga(4)  (Show Source):
You can put this solution on YOUR website! What is the least positive integer n such that n! is divisible by 1000?
List down pairs where product divisible by 10 and have unique members(factorial uses unique consecutive integers):
we need at least 3 pairs (1000 = 10^3)
First two pair
1,10
2,5
Then
3,20
We temporarily choose 20, test if there is another pair with a member lower than 20 and unique member, next is
4,15
15 is lower than 20, we choose 15, test if there is any pair with a member lower than 15 and unique members
5,2, same as 2,5 pair
6,5, not unique (we had 5 in 2,5 pair)
7,10 not unique (we had 10 in 1,10 pair)
8,5 not unique (we had 5 in 2,5 pair)
9,10 not unique (we had 10 in 1,10 pair)
10,1 same as 1,10 pair
11,10 not unique (we had 10 in 1,10 pair)
12,5 not unique (we had 5 in 2,5 pair)
13,10 not unique (we had 10 in 1,10 pair)
14,5 not unique (we had 5 in 2,5 pair)
Therefore 15 is the least positive integer n such that n! is divisible by 1000.
The sum of the first five terms of an arithmetic sequence is 90 less than the sum of the next five terms. What is the absolute difference between two consecutive terms of this sequence? Express your answer as a common fraction.
first 5 terms, t1,t2...t5
next 5 terms, t6,t7...t10
using arithmetic sum Sn
Sn = n(t1 + tn)/2
Sum of first 5 terms:
Sn = 5*(t1 + t5)/2
Sum of other 5 terms
Sn = 5*(t6 + t10)/2
using formula for nth term, tn = t1 + d(n-1), where d = difference of consecutive terms
t5 = t1 + d(5-1) = t1 + 4d
t6 = t1 + d(6-1) = t1 + 5d
t10 = t1 + d(10-1) = t1 + 9d
Sum of first 5 terms = Sum of next 5 terms - 90
5(t1 + t5)/2 = 5(t6 + t10)/2 - 90
5(t1 + t1 + 4d)/2 = 5(t1 + 5d + t1 + 9d)/2 - 90
5(2t1 + 4d)/2 = 5(2t1 + 14d)/2 - 90
5t1 + 10d = 5t1 + 35d - 90
90 = 35d - 10d + 5t1 - 5t1
90 = 25d
d = 90/25
d = 18/5
Answer by AnlytcPhil(1807)  (Show Source):
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