a) How many zeros are at the end of the number $A=2^{5} \times 3^{7} \times 5^{7} \times 11^{3}$?

b) How many zeros are at the end of the number $B=1 \times 2 \times 3 \times 4 \times \cdots \times 137$?
Solution

a) To count the number of zeros at the end of a number, it is enough to find out which is the highest power of 10 that is a divisor of that number. On the other hand, since $10=2 \times 5$, it is enough to know the powers of 2 and 5 that appear in the factorization of the number. So we can write:

$$
A=2^{5} \times 3^{7} \times 5^{7} \times 11^{3}=10^{5} \times 3^{7} \times 5^{2} \times 11^{3}
$$

Since the number $3^{7} \times 5^{2} \times 11^{3}$ is not a multiple of 10, it does not end in 0. Therefore, we see that the original number ends in 5 zeros. In fact, we can calculate the value and get 7277242500000.

b) How do we find out which is the highest power of 10 that divides the number $B$? For this, it is enough to count how many times each factor 2 and 5 appears in the factorization of $B$ and the smaller of these quantities gives the number of zeros at the end of the number $B$.

We start by counting the number of factors 2. Since each even number has a factor 2, we have to count the number of even numbers that exist between 1 and 137. And this is easy to count, as the last even number that appears is 136, which is equal to $2 \times 68$. Therefore, we have 68 even numbers and, therefore, 68 factors 2.

However, that is not all. There are even numbers that contribute more than one factor 2 (for example, 4). In fact, all multiples of 4 contribute two factors 2, so we have to count the multiples of 4 as well, which are in quantity $136 \div 4=34$. Here 136 is the last multiple of 4 that appears in the product. Similarly, there are numbers that contribute three factors, such as multiples of $8=2^{3}$; four factors, such as multiples of $16=2^{4}$, and so on with $32=2^{5}, 64=2^{6}$, and $128=2^{7}$. These are the powers of 2 less than 137. Let's then count all these numbers.

The last multiple of 8 before 137 is 136, so there are $136 \div 8=17$ multiples of 8. The last multiple of 16 is 128, so we have $128 \div 16=8$ multiples of 16. Continuing, there are $128 \div 32=4$ multiples of $32, 128 \div 64=2$ multiples of 64, and one single multiple of 128. Therefore, the number of factors 2 that appear in the factorization of the number $B$ is

$$
68+34+17+8+4+2+1=134
$$

Now, we just need to count the number of factors 5. For this, we follow the same process as above. There are $135 \div 5=27$ multiples of $5, 125 \div 25=5$ multiples of 25, and one single multiple of 125. Therefore, the number of factors 5 is

$$
27+5+1=33
$$

Therefore, since the number of factors 5 is less than the number of factors 2, it means that the number $B$ ends in 33 zeros.