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Thanks a lot Brian. I liked the way you discarded E. I was able to do so but struggled a lot while putting a strong alibi to discard E.Brian@VeritasPrep wrote:Hey guys,
I love that first question, and shovan85 I think you did a great job of explaining it! I may just chime in with some strategic advice if that's okay. Here's how I'd break it down:
To compare exponents, you'll need to find similar bases, so my goals will be:
1) Eliminate choices that just don't have a chance so that I minimize the work.
2) For anything that deserves consideration, find a common base to be able to compare.
3) Because it's asking for the greatest value, run it "tournament style" and only compare each remaining answer choice to the current "leader" for the sake of efficiency.
So...
A: 999^12 is close enough to 1000^12 for comparison that you can call it 1000^12 or (10^3)^12 = 10^36.
B: 10^30 is clearly less than 10^36, so 10^36 (A) still leads
C: 777^10 you could call (3/4 * 1000)^10. 1000^10 is the same as (10^3)^10 or 10^30, and since we know that the 3/4 portion will significantly reduce that (3/4^10 will get pretty small), A still leads.
D: -20^24 is going to be positive so you can ignore the negative. 20^24 = 2^24 * 10^24. This one gets interesting, I think, as 10^24 is obviously smaller than 10^36, our current leader, but 2^24 will be pretty big. Here it's helpful to find common, or at least comparable, bases. 2^24 is the same as (2^3)^8, and we'll use that because it's easier to compare 8 and 10 than it is to compare 2 and 10. 8^8 * 10^24 will clearly be less than 10^36, so D is eliminated.
E: sqrt 15 is the same as 15^1/2, so we can express this as (15^1/2)^40, or 15^20. This should pretty clearly be less than 10^36, but for additional review you might call it (3/2 * 10)^20, or 3/2 ^20 * 10^20. We already know that 2^24 wasn't enough to bump D above 10^36, so 3/2^20 obviously won't do it, and we can eliminate E as well.
The key on most exponent problems is to find common bases, and this one is a great example of that. Thanks for posting this one!
I think what you mean to say is that the denominator is a multiple of 11.neerajkumar1_1 wrote:solving your second question...
the pattern of alternating digits is given when digit 11 is in the denominator...
hence n= 11
so i will pick m= 4
IMO: B
Actually No. Whenever you see an alternating pattern after decimal point the denominator will be 11 other multiples of 11 say 22, 33 cannot give the same pattern. Of course I agree that after certain point of time they will follow the pattern but not the whole figures after the decimal will have the same pattern.TheCloakedMonk wrote:I think what you mean to say is that the denominator is a multiple of 11.neerajkumar1_1 wrote:solving your second question...
the pattern of alternating digits is given when digit 11 is in the denominator...
hence n= 11
so i will pick m= 4
IMO: B
How right you are.shovan85 wrote:Actually No. Whenever you see an alternating pattern after decimal point the denominator will be 11 other multiples of 11 say 22, 33 cannot give the same pattern. Of course I agree that after certain point of time they will follow the pattern but not the whole figures after the decimal will have the same pattern.TheCloakedMonk wrote:I think what you mean to say is that the denominator is a multiple of 11.neerajkumar1_1 wrote:solving your second question...
the pattern of alternating digits is given when digit 11 is in the denominator...
hence n= 11
so i will pick m= 4
IMO: B
Say 5/11 = 0.4545454545.....
but 5/22 = 0.227272727....
Hope I got your idea and you got my idea
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