How many zeroes are there at the end of the following product?
1 × 5 × 10 × 15 × 20 × 25 × 30 × 35 × 40 × 45 × 50 × 55 × 60

The correct option is 10.

To find the number of trailing zeroes at the end of a product, we need to count how many factors of 10 can be formed. Since 10 is the product of prime numbers 2 and 5, each pair of (2 × 5) contributes exactly one zero at the end of the product.
Therefore, the total number of trailing zeroes equals the maximum number of (2 × 5) pairs that can be formed from the prime factorization of all numbers in the given product.

The given expression is:

$P=1\×5\×10\×15\×20\×25\×30\×35\×40\×45\×50\×55\×60$

Let us express each term in the product in terms of its prime factors (specifically focusing on 2s and 5s):

• 1 = 1
• 5 = 5¹
• 10 = 2¹ × 5¹
• 15 = 3¹ × 5¹
• 20 = 2² × 5¹
• 25 = 5²
• 30 = 2¹ × 3¹ × 5¹
• 35 = 5¹ × 7¹
• 40 = 2³ × 5¹
• 45 = 3² × 5¹
• 50 = 2¹ × 5²
• 55 = 5¹ × 11¹
• 60 = 2² × 3¹ × 5¹

Now, let us count the total number of exponents of 2 (factors of 2):

The terms containing factors of 2 are 10, 20, 30, 40, 50, and 60.

$\text{Count of 2s}=1+2+1+3+1+2=10$

So, the total exponent of 2 in the product is 2¹⁰.

Next, let us count the total number of exponents of 5 (factors of 5):

The terms containing factors of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60.

$\text{Count of 5s}=1+1+1+1+2+1+1+1+1+2+1+1=14$

So, the total exponent of 5 in the product is 5¹⁴.

Since a zero is produced by a pair of (2 × 5), the number of such pairs is determined by the limiting factor, which is the smaller exponent between 2 and 5:

$\text{Number of zeroes}=min(\text{Count of 2s},\text{Count of 5s})$

$\text{Number of zeroes}=min(10,14)=10$

Therefore, there are 10 zeroes at the end of the given product.