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15 Aug 2006, 16:36
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How many positive one-digit or two-digit integers cannot be written as the sum of difference of two integers of the form 2^(n-1), where n is a positive integer?
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15 Aug 2006, 23:21
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IMO, the answer should be 39.
The question is talking about only positive single and double digit integers. That covers all numbers between 0 and 100 (exclusive of the 2 extremes).
The question also says that the number should be expressed as a sum or difference of 2^(n-1) where n is positive integer. That means a sum or difference of even numbers.
We know even + even = even
and even - even = even
So all even numbers can be expressed by 2^(n-1). So that leaves us with 50 odd numbers.
But 2^0 = 1. And even - odd = odd. Thus, there will be a few odd numbers that can be obtained by the eqn 2^(n-1) - 2^0 and 2^(n-1) + 2^0.
We need to consider only values of n from 2 to 7 (since 2^7 will become 128 and adding or subtracting 1 from this will give us a number that will still be outside the defined range).
When n=2, 2^1 - 1 = 1 and 2^1 + 1 = 3
When n=3, 2^2 - 1 = 3 and 2^2 + 1 = 5
When n=4, 2^3 - 1 = 7 and 2^3 + 1 = 9
When n=5, 2^4 - 1 = 15 and 2^4 + 1 = 17
When n=6, 2^5 - 1 = 31 and 2^5 + 1 = 33
When n=7, 2^6 - 1 = 63 and 2^4 + 1 = 65
Thus, the total odd numbers that can be represented by adding or subtracting 2 expressions of the form 2^(n-1) = 11
Numbers between 0 and 100 that cannot be represented by 2^(n-1) = 50-11 = 39.
Any other answers to this one please ?
The question is talking about only positive single and double digit integers. That covers all numbers between 0 and 100 (exclusive of the 2 extremes).
The question also says that the number should be expressed as a sum or difference of 2^(n-1) where n is positive integer. That means a sum or difference of even numbers.
We know even + even = even
and even - even = even
So all even numbers can be expressed by 2^(n-1). So that leaves us with 50 odd numbers.
But 2^0 = 1. And even - odd = odd. Thus, there will be a few odd numbers that can be obtained by the eqn 2^(n-1) - 2^0 and 2^(n-1) + 2^0.
We need to consider only values of n from 2 to 7 (since 2^7 will become 128 and adding or subtracting 1 from this will give us a number that will still be outside the defined range).
When n=2, 2^1 - 1 = 1 and 2^1 + 1 = 3
When n=3, 2^2 - 1 = 3 and 2^2 + 1 = 5
When n=4, 2^3 - 1 = 7 and 2^3 + 1 = 9
When n=5, 2^4 - 1 = 15 and 2^4 + 1 = 17
When n=6, 2^5 - 1 = 31 and 2^5 + 1 = 33
When n=7, 2^6 - 1 = 63 and 2^4 + 1 = 65
Thus, the total odd numbers that can be represented by adding or subtracting 2 expressions of the form 2^(n-1) = 11
Numbers between 0 and 100 that cannot be represented by 2^(n-1) = 50-11 = 39.
Any other answers to this one please ?
15 Aug 2006, 23:56
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prashrash... can all even # be expressed as sum or difference of 2^(n-1) try 22??
taking a shot....
n = 1 .. 2 ^0 = 1
n =2 .. 2^1 = 2
n=3... 2^2 =4
... n =7 2^6 = 64
Basically we have 7 numbers.
So now starting with the largest number, 64....when we add each of other the numbers we can get 6 new nos...
now take subtraction from 64 , here we get only 5 distinct numbers (64-32 = 32; already included)
so 64 will give us 11 new numbers.
now lets see 32;
subtraction: will give us 4 unique nos - (the first no has been included in the original series ; i.e. 32-16)
addition: will give us 4 unique nos (the first no has been included in the series from 64; i.e. 32+16 =48)
so 32 will give us 4+4nos
after this is a pattern
16 will give us 3+3 = 6 unique nos
8 will give us 2+2 = 4unique nos
4 will give us 1+1 = 2 unique nos
operations with 2 & 1 will not give us any nos not covered above
so total # of numbers = 2+4+6+8+11+7(original)
=38
total no of one digit numbers = 9
total no of two digit numbers = 100-10 = 90
total = 99
so the # of one or two digit numbers that cannot be expressed at as sum or difference of 2^(n-1) is 99-38 = 61
kevin can OA?
taking a shot....
n = 1 .. 2 ^0 = 1
n =2 .. 2^1 = 2
n=3... 2^2 =4
... n =7 2^6 = 64
Basically we have 7 numbers.
So now starting with the largest number, 64....when we add each of other the numbers we can get 6 new nos...
now take subtraction from 64 , here we get only 5 distinct numbers (64-32 = 32; already included)
so 64 will give us 11 new numbers.
now lets see 32;
subtraction: will give us 4 unique nos - (the first no has been included in the original series ; i.e. 32-16)
addition: will give us 4 unique nos (the first no has been included in the series from 64; i.e. 32+16 =48)
so 32 will give us 4+4nos
after this is a pattern
16 will give us 3+3 = 6 unique nos
8 will give us 2+2 = 4unique nos
4 will give us 1+1 = 2 unique nos
operations with 2 & 1 will not give us any nos not covered above
so total # of numbers = 2+4+6+8+11+7(original)
=38
total no of one digit numbers = 9
total no of two digit numbers = 100-10 = 90
total = 99
so the # of one or two digit numbers that cannot be expressed at as sum or difference of 2^(n-1) is 99-38 = 61
kevin can OA?
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Problem Solving (PS) Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
