Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
Find the number of zeros at the end of the result $3 \times 6 \times 9 \times 12 \times 15 \times$ $\times 99 \times 102$.
(a)4
(b)6
(c)7
(d)10
Answer
Answer (as printed): C
Explanation
Let $N=3 \times 6 \times 9 \times 12 \times \ldots \ldots \ldots+102=3^{34} \times(1 \times 2 \times 3$ $\times 4 \times \ldots \ldots \ldots . \times 34)=3^{34} \times 34!$ Clearly, highest power of 2 in 34 ! is much greater than that of 5. So, number of zeros in $N=$ Highest power of 5 in 34! = $\left[\frac{34}{5}\right]+\left[\frac{34}{5^{2}}\right]=6+1=7$.
Explanation as extracted from the printed page; notation may be imperfect.