Question

first 100 multiplies of 10 are multiplied together. the number of zeroes at te end of the product must be

Answer 1

First 100 multiples of 10 = 10, 20, 30, 40.....990, 1000

We have to find out number of zeros in 10x20x30x40x......990x1000, which we can write as

10^100 (1x2x3x4x.......x99x100)

= 10^100(100!)

Now the number of zeros in the non-factorial part i.e. 10^100 = 100

[Since Factors of 10 are 2 and 5. That means each ten will give you one 5. So (10 * 100times) will give you hundred 5s]

And the number of zeroes in the factorial part i.e. 100! = 100/(5^1) + 100/(5^2) = 100/5 + 100/25 = 20 + 4 = 24

[Since Count all the multiples of 5 in here => 5, 10, 15, 20, 25, ......., 100

Now all of multiples of 5 are divisble by 5^1,5^2 (but not 5^3 as 5^3=125>100)]

So the total umber of zeros in the product = Zeros in non-factorial part + zeros in factorial part i.e. 100 + 24 = 124

so,124 zeros are there

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Answer 2

Answer:- 124

Step-by-step explanation:

Given products = 10×20×30×..........1000

So that..

Required number of zero =( no.of multiples of 5^3)×3+[( no.of multiples of 5^2)-(no of multiples of 25 or 125)]×1.

=(4×3)+(20-4)×2+(100-20)×1

= 12+32+80

=124 answer✓✓