IAS 2020 — CSAT Maths

Correct Option (A)

To determine the number of zeroes at the end of a product, it is necessary to count the total number of prime factors 2 and 5 present in its prime factorization. Each pair of (2 × 5) contributes one zero. The number of zeroes will be equal to the minimum count between the total factors of 2 and the total factors of 5.

The given product is: $1 \times 5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50 \times 55 \times 60$1×5×10×15×20×25×30×35×40×45×50×55×60

Let's list the prime factors of 2 and 5 for each number in the product:

• 5: $5^1$51 (one 5)
• 10: $2^1 \times 5^1$21×51 (one 2, one 5)
• 15: $3 \times 5^1$3×51 (one 5)
• 20: $2^2 \times 5^1$22×51 (two 2s, one 5)
• 25: $5^2$52 (two 5s)
• 30: $2^1 \times 3 \times 5^1$21×3×51 (one 2, one 5)
• 35: $5^1 \times 7$51×7 (one 5)
• 40: $2^3 \times 5^1$23×51 (three 2s, one 5)
• 45: $3^2 \times 5^1$32×51 (one 5)
• 50: $2^1 \times 5^2$21×52 (one 2, two 5s)
• 55: $5^1 \times 11$51×11 (one 5)
• 60: $2^2 \times 3 \times 5^1$22×3×51 (two 2s, one 5)

Total count of prime factor 5:

$1 + 1 + 1 + 1 + 2 + 1 + 1 + 1 + 1 + 2 + 1 + 1 = 14$1+1+1+1+2+1+1+1+1+2+1+1=14

Total count of prime factor 2:

$1 + 2 + 1 + 3 + 1 + 2 = 10$1+2+1+3+1+2=10

The number of zeroes at the end of the product is the minimum of the total counts of prime factors 2 and 5.

Minimum (10, 14) = 10.

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

Incorrect Options:

• Option B (12): This count would be obtained if there were 12 pairs of (2 × 5) factors. This could happen if either the count of factors of 2 or 5 was incorrectly determined to be 12, or if the minimum of the two counts was erroneously calculated as 12. For instance, if one only counted the factors of 5 from numbers ending in 5 or 0, but missed some factors of 2, or miscounted the factors of 5 from 25 and 50.
• Option C (14): This value corresponds to the total number of prime factors 5 present in the product. However, the number of zeroes is limited by the lesser count of either prime factor 2 or 5. Since there are only 10 factors of 2, 14 zeroes cannot be formed.
• Option D (15): This count is incorrect as it overestimates the number of pairs of (2 × 5) factors. A calculation leading to 15 zeroes would imply a significant miscount of either the factors of 2 or 5, or both, exceeding their actual minimum availability.