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Which is greater 105^19 or 100^20?

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Which is greater 105^19 or 100^20? [#permalink] New post 11 Jan 2013, 02:13
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Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.
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Last edited by Orko on 16 Jan 2013, 06:46, edited 1 time in total.
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Re: Question [#permalink] New post 11 Jan 2013, 03:01
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\(100^{20}\) = \(100^{19} * 100\) i.e 100 times \(100^{19}\)

= \(100^{19} + 100^{19} + 100^{19} +\) ...... and so on till hundred times --> (1)

now,

\(105^{19}\) = \((100 + 5) ^{19}\)

Now, as we know \((a + b)^n = a^{19} + ^nC1*a^{(n-1)}*b^1 + ^nC2*a^{(n-2)}*b^2 +\) ........ \(+ ^nCn-1*a^1*b^{(n-1)} + b^n\)

\((100 + 5) ^{19} = 100^{19} + 19*100^{18}*5^1 + 171*100^{17}*5^2 + .... + 5^{19}\)

\(= 100^{19} + 85*100^{18} +\) ... preceding all the terms will be \(< 100^{19}\) --> (2)

so compare (1) and (2) now, we get

\(100^{20} > 105^{19}\)
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Last edited by goutamread on 12 Jan 2013, 00:04, edited 3 times in total.
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Re: Question [#permalink] New post 11 Jan 2013, 03:38
goutamread wrote:
100^20 = 100^19 * 100 i.e 100 times 100^19
= 100^19 + 100^19 + 100^19 + ...... and so on till hundred times --> (1)

now,

105^19 = (100 + 5) ^19
Now, as we know (a + b)^n = a^19 + (nC1)(a^(n-1))(b^1) + (nC2)(a^(n-2))(b^2) + .... + (nCn-1)(a^1)(b^(n-1)) + b^n
(100 + 5) ^19 = 100^19 + (19)(100^18)(5^1) + (171)(100^17)(5^2) + .... + 5^19
= 100^19 + (85)(100^18) + ... preceding all the terms will < 100^19 --> (2)

so compare (1) and (2) now, we get

100^20 > 105^19


Hi Goutamread,

Thanks for the response and effort! Although it seems that you have tried to explain in the best possible way that you are aware of, the explanation has just gone over my head :roll:
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Which is greater 105^19 or 100^20? [#permalink] New post 11 Jan 2013, 03:56
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Let me try :


1st expression : \(105^{19}\) \(= (100 *1.05)^{19}\) \(= 100^{19}*1.05^{19}\) --- [ as \((a*b)^n = (a^n)*(b^n)\) ]

2nd expression : \(100^{20}\) \(= 100^{19} * 100^1\) \(= 100^{19}*100\) --- [ as \(a^n = (a^1)*(a^{n-1})\) ]

now as \(100^{19}\) is common,

Compare \(1.05^{19}\) with \(100\).

as 1.05 is slightly greater than 1

surely, \(1.05^{19} < 100\)


thus

\(105^{19} < 100^{20}\)

Let me know if this worked for you. :!:
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Last edited by goutamread on 12 Jan 2013, 00:20, edited 2 times in total.
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Re: Question [#permalink] New post 11 Jan 2013, 21:05
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ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.


Could you change the main subject as 'Which is greater 105^19 or 100^20?' :) So, It may be helpful for others too, in case they need to dive into the problem.
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P.S. +Kudos Please! in case you like my post. :)

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Re: Question [#permalink] New post 11 Jan 2013, 21:18
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ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.


goutamread has given you a great approach. I would use a similar one though I just try to simplify first. It helps you think faster in some cases.

\(105^{19}\) and \(100^{20}\) have \(5^{19}\) common. Cancel out the common \(5^{19}\) and you are left with

\(105^{19} = 5^{19}*21^{19}\)
and \(100^{20} = 5^{19}*20^{19}*100\)

Now compare \(21^{19}\) and \(20^{19}*100\)

It is easy to make some sense out of \(20^{19}*100\). It is 100 times \(20^{19}\) i.e. you can get this expression if you add 100 terms of \(20^{19}\) together

\(20^{19}*100 = 20^{19} + 20^{19} + 20^{19}\) ... (100 terms) ..............(I)

To compare, we can use binomial for \(21^{19}\). If you do not know how to use binomial, check out this link:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

\(21^{19} = (20 + 1)^{19}\)

Let's try to expand \((20 + 1)^{19}\) using binomial. We will get 19+1 = 20 terms when we expand.

\((20 + 1)^{19} = 20^{19} + 19*20^{18} + 19* 18/2 * 20^{17} + ... + 1^{19}\) ............... (II)

Notice that this expression has 20 terms and none of the terms will be greater than \(20^{19}\).

Compare (I) with (II).
(I) has 100 terms, all of them \(20^{19}\)
(II) has 20 terms, all of them equal to or less than \(20^{19}\).

Hence, (I) will be greater than (II). Or we can say that \(100^{20}\) will be greater than \(105^{19}\).
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Re: Which is greater 105^19 or 100^20? [#permalink] New post 16 Jan 2013, 06:46
goutamread wrote:
Let me try :


1st expression : \(105^{19}\) \(= (100 *1.05)^{19}\) \(= 100^{19}*1.05^{19}\) --- [ as \((a*b)^n = (a^n)*(b^n)\) ]

2nd expression : \(100^{20}\) \(= 100^{19} * 100^1\) \(= 100^{19}*100\) --- [ as \(a^n = (a^1)*(a^{n-1})\) ]

now as \(100^{19}\) is common,

Compare \(1.05^{19}\) with \(100\).

as 1.05 is slightly greater than 1

surely, \(1.05^{19} < 100\)


thus

\(105^{19} < 100^{20}\)

Let me know if this worked for you. :!:


Thanks a lot goutamread! This one is really better. But, I want to know that what made you to split 105 into 100*1.05. This is some thing that is really very important to know because I believe that these are the tactics that one has to think of while attacking a question.
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Re: Question [#permalink] New post 16 Jan 2013, 06:47
goutamread wrote:
ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.


Could you change the main subject as 'Which is greater 105^19 or 100^20?' :) So, It may be helpful for others too, in case they need to dive into the problem.


Done! :-)
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Re: Question [#permalink] New post 16 Jan 2013, 06:50
VeritasPrepKarishma wrote:
ObsessedWithGMAT wrote:
Dear fellowmembers,

I have come across a question that I am not able to either solve it or find a suitable explanation; the question is: -

Q). Which is greater 105^19 or 100^20?

Please help!
Thanks in advance.


goutamread has given you a great approach. I would use a similar one though I just try to simplify first. It helps you think faster in some cases.

\(105^{19}\) and \(100^{20}\) have \(5^{19}\) common. Cancel out the common \(5^{19}\) and you are left with

\(105^{19} = 5^{19}*21^{19}\)
and \(100^{20} = 5^{19}*20^{19}*100\)

Now compare \(21^{19}\) and \(20^{19}*100\)

It is easy to make some sense out of \(20^{19}*100\). It is 100 times \(20^{19}\) i.e. you can get this expression if you add 100 terms of \(20^{19}\) together

\(20^{19}*100 = 20^{19} + 20^{19} + 20^{19}\) ... (100 terms) ..............(I)

To compare, we can use binomial for \(21^{19}\). If you do not know how to use binomial, check out this link:
http://www.veritasprep.com/blog/2011/05 ... ek-in-you/

\(21^{19} = (20 + 1)^{19}\)

Let's try to expand \((20 + 1)^{19}\) using binomial. We will get 19+1 = 20 terms when we expand.

\((20 + 1)^{19} = 20^{19} + 19*20^{18} + 19* 18/2 * 20^{17} + ... + 1^{19}\) ............... (II)

Notice that this expression has 20 terms and none of the terms will be greater than \(20^{19}\).

Compare (I) with (II).
(I) has 100 terms, all of them \(20^{19}\)
(II) has 20 terms, all of them equal to or less than \(20^{19}\).

Hence, (I) will be greater than (II). Or we can say that \(100^{20}\) will be greater than \(105^{19}\).


Dear Karishma,

Thanks a lot for taking out time to reply to my question! I am still struggling with the fact that this question is from Inequalities and neither you nor goutamread applied any concept of Inequalities or is it that I am not able to identify?
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Re: Which is greater 105^19 or 100^20? [#permalink] New post 17 Jan 2013, 10:52
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Here is the smart solution. Lets assume both are equal. If after some manipulation we can show LHS > RHS or LHS < RHS we would have got the answer.

105^19 = 100^20

Take 100^19 to the LHS (Divide both sides by 100^19)

We have 1.05^19 = 100

From Binomial theorem we know than (1+x)^n = 1 + nx when x<<1

Hence 1.05^19 = (1 + 0.05*19) = 1.95 < 100

Hence 105^19 < 100^20
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Re: Which is greater 105^19 or 100^20? [#permalink] New post 18 Jan 2013, 07:50
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ObsessedWithGMAT wrote:
Thanks a lot goutamread! This one is really better. But, I want to know that what made you to split 105 into 100*1.05. This is some thing that is really very important to know because I believe that these are the tactics that one has to think of while attacking a question.



I'm not an expert, rather I'm just another aspirant on the forum. I could try to explain:

See, whenever we deal with a big number/expression, we try to simplify the expression to make it realistic :P

There could be numerous ways to solve a question, what matters is which one you feel at your ease... now as you mentioned inequality,

assume \(105^{19} > 100^{20}\) lets try to prove whther our assumption is correct or not. If yes, then \(105^{19} > 100^{20}\) and if not then, \(105^{19} < 100^{20}\)

\(105^{19} > 100^{20}\)
Now, don't these expressions scare us. yes they are scary.. so lets try and simplify.
lets divide both side by \(100^{20}\), (as \(100^{20}\) is +ve --> inequality sign wont change)
\(\frac{105^{19}}{100^{20}} > 1\)
\(\frac{105^{19}}{{100^{19}*100}} > 1\)
\((\frac{105}{100})^{19}*\frac{1}{100} > 1\)
\(\frac{(1.05) ^{19}}{100} > 1\) OR \((\frac{26}{25}) ^{19}* \frac{1}{100} > 1\) ==> \(26 ^{19} > 100 * 25^{19}\)

From here its already discussed earlier how to solve the problem. Take away could be : There are various ways to solve a problem, what matters is what suits you, but you should be aware of more than one trick.
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Last edited by goutamread on 20 Jan 2013, 02:23, edited 1 time in total.
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Re: Which is greater 105^19 or 100^20? [#permalink] New post 18 Jan 2013, 08:39
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For more practice : which-is-greater-54-200-or-5145.html
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Re: Which is greater 105^19 or 100^20?   [#permalink] 18 Jan 2013, 08:39
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