Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
The numbers 2, 4, 6, 8, $\_\_\_\_$ 98, 100 are multiplied together. The number of zeros at the end of the product must be
(a)10
(b)11
(c)12
(d)13
Answer
Answer (as printed): C
Explanation
$$\begin{aligned} N & =2 \times 4 \times 6 \times 8 \times \ldots \ldots \ldots \times 98 \times 100 \\ & =2^{50} \times(1 \times 2 \times 3 \times \ldots \ldots \ldots \times 49 \times 50)=2^{50} \times 50! \end{aligned}$$ Clearly, the highest power of 2 in $N$ is much higher than that of 5. ∴ Number of zeros in $N=$ Highest power of 5 in $N=$ $\left[\frac{50}{5}\right]+\left[\frac{50}{5^{2}}\right]=10+2=12$.
Explanation as extracted from the printed page; notation may be imperfect.