Find the number of zeroes in the product :(1raised to power 1)*(2 raised power 2)*(3 raised power 3)*...(100?

Question

Find the number of zeroes in the product :(1 raised to power 1)*(2 raised power 2)*(3 raised power 3)*...(100 raised to power 100).
(a) 1200
(b) 1300
(c) 1050
(d) None of these

Answer 1

After staring for a while, I finally decided that the previous respondant was trying to use the formula for the sum of the first n numbers of the series a + 2a + 3a + ... + na, and then didn't know what to do with the answer. Maybe I am wrong, but he surely needs to explain his notation before trying to give an intelligible answer. In any case, all of the previous respondants are wrong, the answer is (b) 1300.

To see why, notice that the number of zeros at the end of the product will be the number of 10's dividing that number, and this, in turn, is the minimum of the number of 5's dividing the number and the number of 2's dividing the number, since we need to pair off a factor of 2 and a factor of 5 to form a factor of 10. Clearly, there are more 2's dividing the number, so we just need to know the number of 5's to find the number of zeros at the end of its decimal representation.

To count the number of 5's dividing the number, notice that there will only be 5's appearing in the terms 5^5 * 10^10 * 15^15 ... Notice that the first term here contributes 5 factors of 5, the second contributes 10 factors of 5, the third contributes 15 factors of 5, etc. So at a first approximation, we might guess that the number of factors of 5 in the number given in the problem is

5 + 10 + 15 + ... + 100 = 5(1 + 2 + 3 + ... + 20).

The well-known formula:

1 + 2 + ... + n = n(n+1)/2

shows that the above sum is
5 + 10 + 15 + ... + 100 = 5(20*21/2) = 1050.

If you don't know the "well-known" formula, it can be shown by an easy induction. Otherwise, look up a biography of Gauss -- all of them mention the story of how he rediscovered this ancient formula by himself when he was 10 years old.

In any case, you might think that 1050 is the answer, but in fact, we have missed some factors of 5! Notice that in the terms 25^25, 50^50, 75^75 and 100^100, in the above computation of counting the number of 5's, we only counted 25, 50, 75 and 100 5's as being contributed by each of these factors. However, these factors are powers of 5^2 raised to a power, so in fact, 25^25 = 5^(50) contributes 50 factors, and similarly, 50, 75, and 100 contribute 100, 150, and 200 factors of 5 respectively to the product. Since we have only counted them as contributing 25, 50, 75, and 100 so far, we need to add another 25, 50, 75, and 100 to our count.

Thus, the answer to the question is our previous 1050 plus 25+50+75+100, so we get

1050 + 25 + 50 + 75 + 100 = 1350.

Thus, the answer is (b).

Answer 2

The answer I think is 1050
since...

(A sub 1) = 5
(A sub n) = 95
n = ?
d = 10

(A sub n) = (A sub 1) + (n - 1)d
95 = 5+(n-1)10
90 = (n-1)10
9 = n-1
n = 10

(S sub n) = (n/2)[(a sub 1) + (a sub n)]
= (10/2)(5+95)
= 5(100)
= 500

next...

(S sub n) = (10/2)(10+100)
= 5(110)
= 550

and...
500 + 550 = 1050

I did it like this since if you multiply a 5 with an even number you get a zero and the first one would be all the fives the second one would be all the tens since a ten already has a zero

Answer 3

surely (d)

Why not do 100! and square it?

Answer 4

the answer is c --1050

don't ask me how but calculate for answer

Answer 5

none of these