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How many different positive numbers smaller than 2*10^8 can be formed

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How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post 13 Mar 2016, 05:50
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How many different positive numbers smaller than \(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561
[Reveal] Spoiler: OA

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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post 13 Mar 2016, 05:59
chetan2u wrote:
How many different positive numbers smaller than\(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561



OA after 3 days


Any combination of number starts with digit "2" is greater than 2*10^8
Total possible combinations with 1 and 2 = 2^9
Total possible combinations with 2 as starting digit = 2^8

Numbers smaller than 2*10^8 can be formed = 2^9 - 2^8
= 2^8(2-1)
= 256.

IMO A

Another method:

Start with most restrictive clause:

Any number starts with 1 is smaller than 2*10^8.

1*2^8
=256

IMO A
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Re: How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post Updated on: 13 Mar 2016, 15:00
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chetan2u wrote:
How many different positive numbers smaller than\(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561


Let's start with all 9-digit numbers that are less than 200,000,000
So, let's take the task of creating 9-digit numbers and break it into stages.

Stage 1: Select the 1st digit.
This digit must be 1, so we can complete stage 1 in 1 way

Stage 2: Select the 2nd digit.
This digit can be either 1 or 2, so we can complete stage 2 in 2 ways

Stage 3: Select the 3rd digit.
This digit can be either 1 or 2, so we can complete stage 3 in 2 ways

Stage 4: Select the 4th digit.
This digit can be either 1 or 2, so we can complete stage 4 in 2 ways

.
.
.
.
.
Stage 9: Select the 9th digit.
This digit can be either 1 or 2, so we can complete stage 3 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 9 stages (and thus create a 9-digit number) in (1)(2)(2)(2)(2)(2)(2)(2)(2) ways
This equals 2^8.



Now, we'll count all 8-digit numbers
So, let's take the task of creating 8-digit numbers and break it into stages.

Stage 1: Select the 1st digit.
This digit can be 1 or 2 , so we can complete stage 1 in 2 ways

Stage 2: Select the 2nd digit.
This digit can be either 1 or 2, so we can complete stage 2 in 2 ways
.
.
.
.
.
[u]Stage [8/u]: Select the 8th digit.
This digit can be either 1 or 2, so we can complete stage 3 in 2 ways

By the FCP, the total number of outcomes = 2^8


If we follow the same steps for 7-digit numbers, we get a total of 2^7 outcomes.

And so on.

So, the TOTAL number of possibilities = 2^8 + 2^8 + 2^7 + 2^6 ..... + 2^1 = 766

Answer: D
--------------------------

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. For more information about the FCP, watch our free video: http://www.gmatprepnow.com/module/gmat-counting/video/775

You can also watch a demonstration of the FCP in action: https://www.gmatprepnow.com/module/gmat ... /video/776

Cheers,
Brent
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Originally posted by GMATPrepNow on 13 Mar 2016, 11:29.
Last edited by GMATPrepNow on 13 Mar 2016, 15:00, edited 1 time in total.
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Re: How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post 13 Mar 2016, 14:34
chetan2u wrote:
How many different positive numbers smaller than \(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561



OA after 3 days


So we are looking for ALL the numbers less than \(2*10^8\).

Start with single-digit numbers. We only have 1 and 2.

Move on to two-digit numbers. There, we have four possible numbers - 11, 12, 21, 22.

At this point, a trend should be becoming noticeable here. The number of ways you can form n-digit numbers using only the numbers 1 and 2 is \(2^n\). This is pretty sensible from a permutation perspective, given that each digit in any n-digit number can be thought of as slots in which you are allowed to place any combination of 1s and 2s.

Now we are working with numbers that can grow to be nine-digits long, but we have to mindful of the ceiling the problem places on us - \(2*10^8\). The number of nine-digit numbers that we can form are limited to those numbers that have 1 as its first digit. The rest of the 8 digits can be populated with any combination of 1s and 2s.

Therefore, for nine-digit numbers, we have \(2^8\) different numbers that can be formed using only 1s and 2s. For eight-digit numbers, we also have \(2^8\) different numbers that can be formed (the first digit is not constrained to 1). For seven-digit numbers, we have \(2^7\) different possibilities. For six-digit numbers, we have \(2^6\) possibilities. A pattern should be becoming evident here. Once you see that pattern, you can form an equation that will give you the answer.

\(2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1\)

\(256 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 = 766\)

The answer is 766 (D)
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Re: How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post 14 Mar 2016, 03:27
sowragu wrote:
chetan2u wrote:
How many different positive numbers smaller than\(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561



OA after 3 days


Any combination of number starts with digit "2" is greater than 2*10^8
Total possible combinations with 1 and 2 = 2^9
Total possible combinations with 2 as starting digit = 2^8

Numbers smaller than 2*10^8 can be formed = 2^9 - 2^8
= 2^8(2-1)
= 256.

IMO A

Another method:

Start with most restrictive clause:

Any number starts with 1 is smaller than 2*10^8.

1*2^8
=256

IMO A


Hi,
you have understood the Q well but made a critical error..

The Q asks for all smaller numbers but you have found only for 9-digits number..
your Solution would have been OK had the Q been
How many different 9-digit positive numbers smaller than\(2*10^8\) can be formed using the digits 1 and 2 only?

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Expert Post
Math Expert
User avatar
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Joined: 02 Aug 2009
Posts: 5777
Re: How many different positive numbers smaller than 2*10^8 can be formed [#permalink]

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New post 28 Dec 2017, 05:54
chetan2u wrote:
How many different positive numbers smaller than \(2*10^8\) can be formed using the digits 1 and 2 only?

A. 256
B. 510
C. 512
D. 766
E. 6561



OA after 3 days


Hi,
such Q requires initially understanding the Problem in terms of number and then Progressions-

since here we have 1 digit number to 9 digit number, 2*10^8..
so lets work out digit wise-
1) 1-digit-- 1 and 2 : total- \(2^1\)way
2) 2-digit-- two places and can be filled by two numbers 1,2 : ways= \(2*2=2^2\)
3) 3-digit-- 3 places and can be filled by two numbers 1,2 : ways=\(2*2*2=2^3\)
and so on till ..
8) 8-digit-- \(2^8\)
9) 9-digit-- \(2^9\).. but here only numbers with first digit as 1 as we require number\(<2*10^8\) : total\(\frac{2^9}{2}\)

Total ways= \(2^1+2^2+2^3+2^4+2^5+2^6+2^7+2^8+\frac{2^9}{2}\)
this can also be solved with Geometric Progression
SUM= \(\frac{a(r^N-1)}{(r-1)}\).. where r=2, a=2^1, N=8..
so\(total = \frac{2(2^8-1)}{(2-1)} + \frac{2^9}{2} =2*255 + \frac{512}{2}=510+256=766\)
Ans= 766
D

_________________

Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


GMAT online Tutor

Re: How many different positive numbers smaller than 2*10^8 can be formed   [#permalink] 28 Dec 2017, 05:54
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