The numbers 1, 2, 3, 4, ......, 1000 are multiplied together. The number of zeros at the end (on the right) of the product must be :
A. 30
B. 200
C. 211
D. 249
Answer: Option D
Solution(By Examveda Team)
Let N = 1 × 2 × 3 × 4 × ..... × 1000 = 1000!Clearly, the highest power of 2 in N very high as compared to that of 5.
So, the number of zeros in N will be equal to the highest power of 5 in N.
∴ Required number of zeros
= $$\left[ {\frac{{1000}}{5}} \right] + \left[ {\frac{{1000}}{{{5^2}}}} \right] + \left[ {\frac{{1000}}{{{5^3}}}} \right]$$ $$ + \left[ {\frac{{1000}}{{{5^4}}}} \right]$$
= 200 + 40 + 8 + 1
= 249
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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