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Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.
The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.
Statement 2 tells us that d > 6, which is a worthless piece of information on its own.
When we combine the 2 statements, we get C.
There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.
Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.
The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.
Statement 2 tells us that d > 6, which is a worthless piece of information on its own.
When we combine the 2 statements, we get C.
There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.
i didnt get how we get "total of seven 5s"..i am able to see only six 5's.
Statement 1 tells us that we need to find out how many times is 30! divisible by 10. The hardest way to solve this is to break down 30! to its prime factors and count the 2s and 5s, because they make up the 10s. It is pretty easy to see that there are many more 2s than 5s in 30!, because we have 15 even numbers and only 6 numbers divisible by 5.
The numbers that contain 5s are 5=5, 2*5=10, 3*5=15, 4*5=20, 5*5=25, 6*5=30. So we have a total of seven 5s and more than seven 2s, which means that 30! can be evenly divided by 10 up to seven times. Therefore 1 <= d <=7. We can't figure out the exact value, so the statement is insufficient.
Statement 2 tells us that d > 6, which is a worthless piece of information on its own.
When we combine the 2 statements, we get C.
There was a very nice discussion of a similar problem about a month ago, but I can't find the post. The approach is "stolen" from there.
i didnt get how we get "total of seven 5s"..i am able to see only six 5's.
I think it is C
F = 1*2*3...*30
From A we know 10^d * X = F means..
and F contains for 7 instances of (5*2)
as in... 1*2*3*4...10 has two (5*2 and 10)
11 to 20 has (15 and 20)(meaning another 5 and 2*10 makes 2)
21 to 30 has a 25 and 30 (5 * 5 = 25 and 3 *10...for makes 3 instances of 10)
so d could be from 1 to 7...
from statement 2 u get that d > 6
Thus combining both u get the exact vlue of d...
Hence C
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