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This question belongs to Arithmetic Ability Number System
Number System
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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 :

Answer & Solution
Correct Answer: Option D
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
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