We'll know the number of ending zeros of 100! by finding the largest power of 10 contained in 100!
Why? Because for example 1000= 10^3;so 1000 has 3 ending zeros.
900= 9*10^2; so 900 has 2 ending zeros.
Now to solve the problem, we first need to break down 10 into prime numbers. 10 = 2*5.
We then need to find how many 2s (call it x) and how many 5s (call it y) go in 100!
Once we do that, we'll take the lower number between x and y.
Clearly, there will be more 2s than 5s in 100!; so let's save time by focusing on y (the number of 5).
Let's do it:
Divide 100 by 5 and the resulting quotient by 5 repeatedly until the quotient of the division is less than 5, which is the divisor, and you stop.
We only write out and take the quotients in the divisions.
100/5 = 20; 20/5= 4. Stop! since 4 is less than 5.
Let's now add all the quotients: y= 20 + 4 = 24
24 is the largest power of 5 in 100!24
is also the largest power of 10 contained in 100!100! has 24 ending zeros!
That's all folks!
Asan Azu, The GMAT Doctor.
Asan is a very experienced GMAT teacher and tutor that can be reached at http://www.GMATLounge.com
For 4500, the same methodology applies. Remember that we only take the quotients in the successive divisions by 5.
4500/5 = 900; 900/5 = 180; 180/5 = 36; 36/5 = 7; 7/5 = 1; Stop! since 1 is less than 5.
Now add the quotients : y= 900+180+36+7+7 = 1124.1124
is the largest power of 10 contained in 4500!4500! has 1124 ending zeros.
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Dakar Azu is The GMAT Doctor.
Dakar is an experienced GMAT teacher who can be reached at http://700gmatclub.com. He prepares aspiring business students thoroughly to get them well over the GMAT 700-score hurdle through his online GMAT courses.