1.Statement 1 gives you a clue that 10^d means that the product of 1*2*...*30 will end in a certain number of 0's. How many, you ask? Well, lets see:
You can write out the product entirely and see for yourself: 1*2*3*4*5*6*7*8*9*10*11*12*13*14*15*16*17*18*19*20*21*22*23*24*25*26*27*28*29*30
Automatically, pull out 10,20, and 30, since they end in 0. 10*20*30 = 6000, which gives you 3 zeros. Now, you have to remember that the only other number when multiplied that can yield a zero is 5. So you have 5, 15, and 25 left. Now find how many zeros you get when you multiply any of those numbers ending in 5:
25 *4 = 100, which gives you 2 zeros.
15 * 6 = 90, which gives you 1 zero
5 * 2 = 10, which gives you 1 zero.
You're total is 3 + 2 + 1 + 1 = 7. Choice C is your answer.
------------------------------------------
2.Not sure about notation. Sorry
------------------------------------------
3. Use numbers:
Make 2 sets is as follows {1,2,70,127,130}, {69,69,70,71,71} which satisfies the conditions
In both cases, the average of the 2 greatest integers are > median and average of the set. Always better to use numbers as examples here.
------------------------------------------
4. Average # of books per student = 2, there are 30 students total, so total # of books = 60.
12 students borrowed 1 book, and 10 borrowed 2 books. That means that 12*1 + 10*2 = 32 books were used out of the 60, so there are 60 - 32 = 28 books remaining.
Those 28 books remain left for 6 other students (since there are 30 students, 2 which borrow nothing, leaving 28 students to borrow entirely, 12 borrow 1 book and 10 borrow 2 books, which leaves 6 students left who borrow at least 3).
So if we had 5 students who borrowed only 3 (maximizing the number of books the 6th person could take), then that means you would have 28 - (5 * 3) = 13 books left for the 6th student to take.
