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Solving expression

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Solving expression

by nasir » Thu Sep 30, 2010 8:04 am
Which of the following expressions has the greatest value?



* 999^{12}

* 10^{30}

* 777^{10}

* (-20)^{24}

* sqrt{15})^{40}


mand n are integers. What is the smallest possible value of integer m if m/n = 0.3636363636...?



3
4
7
13
22
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by neerajkumar1_1 » Thu Sep 30, 2010 8:49 am
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
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by shovan85 » Thu Sep 30, 2010 9:21 am
1. IMO A

The explanation to this problem is very tedious. Still I will try my best.

The below method what I have shown is not at all advisable. These calculations should be intuitive.

A Vs B --> 999^12 Vs 10^30: we can say 999 nearly 1000 = 10^3 thus 10^36 > 10^30, proceed with A

A VS C --> 999^12 > 777^10 clearly proceed with A

A VS D --> 999^12 Vs (-20)^24
(-20)^24 = (2^24)*(10^24)
divide 10^36 (from A) by [(2^24)*(10^24)]
= (10^36)/[(2^24)*(10^24)]
= (10^12)/[(2^12)*(2^12)]
= 5^12 / 2^12 Now clearly 5^12 > 2^12 so proceed with A.

A Vs E -->E can be said 15^20= (3^20) * (5^20)
999^12 = (3^24) * (111^12)
Now if u see 3^24 > 3^20 so leave this part.
consider 5^20 = 25^10 so what can u see now 25^10 < 111^12. Hence A is the answer.
:cry:

2 is explained well in the above post.[/b]
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by nasir » Fri Oct 01, 2010 8:33 am
THanks guys. Both answers are right
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by Brian@VeritasPrep » Fri Oct 01, 2010 9:42 am
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!
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by shovan85 » Fri Oct 01, 2010 10:08 am
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!
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.
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by mrinal2100 » Fri Oct 01, 2010 11:37 am
while i was solving this question i approached it in different way.Don't know if its right or wrong.

like for A when we multiply a 3 digit no with another 3 digit the result is 5 digit no.Now when this 5 digit no is again multiplied with 3 digit number it results in 7 digit number.
As a result a pattern is formed when we keep on multiplying the result with 3 digit no.

the resulting digits would be 5,7,9,11,13,15,17,19,21,23,25 digit no for option A since 999 is being multiplied 12 times.

like this we can see that when a 2 digit no is multiplied by itself it results in 3 digit no and when 3 digit result is again multiplied with the 2 digit no it results in 4 digit no....

so for 2 digit no the pattern is 3,4,5,6,7,8,9......

in this way we can compare the final answer based on the no of digits.
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by TheCloakedMonk » Fri Oct 01, 2010 11:49 am
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
I think what you mean to say is that the denominator is a multiple of 11.
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by shovan85 » Fri Oct 01, 2010 12:04 pm
TheCloakedMonk wrote:
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
I think what you mean to say is that the denominator is a multiple of 11.
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.

Say 5/11 = 0.4545454545.....
but 5/22 = 0.227272727....

Hope I got your idea and you got my idea :)
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by TheCloakedMonk » Fri Oct 01, 2010 1:14 pm
shovan85 wrote:
TheCloakedMonk wrote:
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
I think what you mean to say is that the denominator is a multiple of 11.
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.

Say 5/11 = 0.4545454545.....
but 5/22 = 0.227272727....

Hope I got your idea and you got my idea :)
How right you are.

I meant 11, 111, 1111, etc...

Sorry for the miscommunication. That was sloppy of me.
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by shovan85 » Fri Oct 01, 2010 8:15 pm
TheCloakedMonk wrote:
How right you are.

I meant 11, 111, 1111, etc...

Sorry for the miscommunication. That was sloppy of me.
Oh!! then its correct. Thanks :)
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