Hi sukhman,
For future reference, it's best to post ONLY 1 question per thread. This will keep each "conversation" focused on one subject and cut down on any confusion.
Since all of these questions deal with 127!, the basic approach will be the same: find the "numbers" you're looking for.
127! = (127)(126)(125)......(3)(2)(1)
Question 1: To find the highest "power of 5", we need to find "all the 5s" in this string of numbers.
There are 25 multiples of 5 = 25 5s
HOWEVER, there are more than just those 25!!!
The following numbers have "extra" 5s:
25 = one extra 5
50 = one extra 5
75 = one extra 5
100 = one extra 5
125 = TWO extra 5s
So, there are 25 + 6 = 31 5s; the highest power is 5^31.
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Question 2: 58^n divides into 127!; what is the maximum value of n?
58 = 29x2
There are lots of 2s in 127!, so we need to find all of the 29s in 127!
The following numbers all contain one 29:
29
58
87
116
So, the maximum n = 4
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Question 3: Find the number of zeroes in 127!
A "zero" requires either a number that "ends in zero" OR "5 times an even"
The following numbers all contain zeroes: 10,20,30,40,50,60,70,80,90,100,110,120
***NOTE: 100 contains TWO zeroes.
Total of 13 zeroes.
The following numbers would have a zero when multiplied by 2: 5,15,25,35,45,55,65,75,85,95,105,115,125
Total of 13 zeroes.
The numbers 25, 75 and 125 share a property worth noting:
25 x 2 x 2 = 100, which is 2 zeroes (not just 1)
75 x 2 x 2 = 300, which is 2 zeroes (not just 1)
125 x 2 x 2 x 2 = 1000, which is 3 zeroes (not just 1)
So we have to factor in the "extra 0s" to the calculation...
Total of ALL zeroes = 13 + 13 + 4 = 30
GMAT assassins aren't born, they're made,
Rich
Last edited by
[email protected] on Mon Oct 14, 2013 2:07 pm, edited 2 times in total.
