What is the number of zeros at the end of the product 5⁵ × 10¹⁰ × 15¹⁵ × … ….. × 125¹²⁵ ? · ABC26GN0039
Subject: General Aptitude · Chapter: Number System
What is the number of zeros at the end of the product $5^{5} \times 10^{10} \times 15^{15} \times \ldots \ldots . . \times 125^{125}$ ?
Answer
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Explanation
Clearly, the highest power of 2 is less than that of 5 in N. So, the highest power of 2 in N shall give us the number of zeros at the end of N. Highest power of 2 = Number of multiples of 2 + Number of multiples of 4 (i.e. 22) + Number of multiples of 8 (i.e. 23) + Number of multiples of 16 (i.e. 24) = [(10 + 20 + 30 +....... + 120) + (20 + 40 + 60 +........ + 120) + (40 + 80 + 120) + 80] = (780 + 420 + 240 + 80) = 1520. Hence, required number of zeros = 1520.
Explanation as extracted from the printed page; notation may be imperfect.