Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
The number of zeros at the end of the product $5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50$ is
(a)5
(b)7
(c)8
(d)10
Answer
Answer (as printed): C
Explanation
Let $N=5 \times 10 \times 15 \times 20 \times 25 \times 30 \times 35 \times 40 \times 45 \times 50$ $$=5^{10} \times(1 \times 2 \times 3 \times 4 \times \ldots \ldots \ldots \times 10)=5^{10} \times 10!$$ Highest power of 2 in $10!=\left[\frac{10}{2}\right]+\left[\frac{10}{2^{2}}\right]+\left[\frac{10}{2^{3}}\right]=5+2+1=8$. Highest power of 5 in $10!=\left[\frac{10}{5}\right]=2$. $$\therefore N=2^{8} \times 5^{12} \times k .$$ Since highest power of 2 is less than that of 5, so required number of zeros $=8$.
Explanation as extracted from the printed page; notation may be imperfect.