Counting Zeros in the Product of Natural Numbers up to 1962!

[Kartik.]

Question- How many zeros are there in the product of all the natural numbers right from 1 to 1962 or in short in 1962! ?

Attempt-
Well,
The product as a product of prime numbers, will denoted as -
Let the product be N,

N = 2^a₁ . 3^a₂ . 5^a₃ . ... p^a_x


As far as i see it we will have to count the number of 5s and 2s that occur in the product N's representation above and also the factors which will yield 5s or 2s.So by far we are concerned only to the numbers a₁ and a₃.And here lies my problem, how will we calculate the number of occurrences and also figure out that how many other factors will yield how many 5s or 2s ?

 

[Office_Shredder]

You only need to calculate a₃, the value of a₁ is not needed for the answer (of course you should figure out why)

To figure out what a₃ is, it might help to first reason how you could calculate the number of times 5 divides a smaller number like 100! (where you can calculate this by hand) and then see how you would extend the result

 

[Infinitum]

The number of 2's is obviously greater than the number of 5's, so there's always a 2 for each 5 to give a 10, and that reduces your work by half Now, the total number of 5's in any given number will be equal to the number of times it can be divided by 5. But, there are some terms which have two 5's, meaning they are multiples of 25. Here you need not count the one's you already got due to division by 5. Next would be 5*5*5, that's 125 and so on. Hope this gives you an idea.

 

[DonAntonio]

Question- How many zeros are there in the product of all the natural numbers right from 1 to 1962 or in short in 1962! ?

Attempt-
Well,
The product as a product of prime numbers, will denoted as -
Let the product be N,

N = 2^a₁ . 3^a₂ . 5^a₃ . ... p^a_x


As far as i see it we will have to count the number of 5s and 2s that occur in the product N's representation above and also the factors which will yield 5s or 2s.So by far we are concerned only to the numbers a₁ and a₃.And here lies my problem, how will we calculate the number of occurrences and also figure out that how many other factors will yield how many 5s or 2s ?

As the number of 2's is clearly way larger than that of 5's, all you've to do is calculate what's the highest power of

5 that divides $\,\,1962!\,\,$, which is given by $$\displaystyle{\sum_{n=1}^\infty\left[\frac{1962}{5^n}\right]}$$ with $\,\,[x]=\,\,$ the integer part of the real number x (why is this formula true and why is the above a finite sum?)


BTW, and if you're interested, the number is $\,\,488 = 392+78+15+3$

DonAntonio

 

[CellCoree]

Well, to find the number of zeros in a product, we need to consider the prime factorization of each number up to 1962 and see how many times the prime factors 2 and 5 occur. Since we are only concerned with 2 and 5, we can ignore all other prime factors.

So, let's start by finding the number of 5s. We know that every multiple of 5 (5, 10, 15, etc.) will contribute at least one 5 to the product. There are 392 multiples of 5 up to 1962, and each of these contributes at least one 5. But we also need to consider the multiples of 25 (25, 50, 75, etc.), which will contribute another 5 to the product. There are 78 multiples of 25 up to 1962. Similarly, we need to consider the multiples of 125 (125, 250, 375, etc.), which will contribute yet another 5 to the product. There are 15 multiples of 125 up to 1962.

So, in total, we have 392 + 78 + 15 = 485 5s in the product.

Now, let's consider the number of 2s. Every multiple of 2 (2, 4, 6, etc.) will contribute at least one 2 to the product. There are 981 multiples of 2 up to 1962, and each of these contributes at least one 2. But we also need to consider the multiples of 4 (4, 8, 12, etc.), which will contribute another 2 to the product. There are 490 multiples of 4 up to 1962. Similarly, we need to consider the multiples of 8 (8, 16, 24, etc.), which will contribute yet another 2 to the product. There are 245 multiples of 8 up to 1962. Continuing this pattern, we also need to consider the multiples of 16, 32, 64, and 128, which will contribute more 2s to the product.

So, in total, we have 981 + 490 + 245 + 122 + 61 + 30 + 15 + 7 + 3 + 1 = 1954 2s in the product.

Since we need both