Subject: General Aptitude · Chapter: Number System · Exam: · Marks: · Difficulty:
The numbers 1, 3, 5, 7, 99 and 128 are multiplied together. The number of zeros at the end of the product must be
(a)Nil
(b)7
(c)19
(d)22
Answer
Answer (as printed): B
Explanation
Let $N=(1 \times 3 \times 5 \times 7 \times \ldots \times 99) \times 128$. Clearly, $N$ contains 10 multiples of 5 (5, 15, 25, 35, ....., 95) and only one multiple of 2 i.e. 128 or $2^{7}$. Clearly, highest power of 5 in $N$ is greater than that of 2. ∴ Number of zeros in $N=$ Highest power of 2 in $N=7$.
Explanation as extracted from the printed page; notation may be imperfect.