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: Option C
Solution(By Examveda Team)
N = 2 × 4 × 6 × 8 × ..... × 98 × 100= 250 (1 × 2 × 3 × ..... × 49 × 50)
= 250 × 50!
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
Related Questions on Number System
Three numbers are in ratio 1 : 2 : 3 and HCF is 12. The numbers are:
A. 12, 24, 36
B. 11, 22, 33
C. 12, 24, 32
D. 5, 10, 15
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