2012 AIME I Problem #5

 

Problem: Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

First note that it is impossible to just have the two numbers end in a 1 and a 0. This is because we have an uneven number of 1's and 0's remaining, and they can not cancel each other out to make 0. Thus, we need to add constraints to these numbers in order to achieve the task. Next note that 10_2-1_2=1_2, so an option is to force the two numbers to end in 10 and 01. This works, because now we have an even number of zeros and ones distributed among the remaining digits of the two binary integers.

Thus, when we place the integers as explained, we have 4 0s and 7 1s to distribute among the remaining 11 digits of either number. (We say either because we we will then place the digits of the other integer to match up with the digits of the first integer) Thus we can say that we are choosing 4 places out of 11 spots to put the four 1's we need to complete the requirements, and we have \displaystyle\binom{11}{4}=\boxed{330} ways to arrange the digits.

An example of two integers that work is

\setlength{\tabcolsep}{0.5mm}\begin{array}{cccccccccccccc}\\&1&0&0&1&0&0&1&0&0&1&...

Notice that 10 and 01 are the last two digits for these integers, and all the other digits match up so that they cancel when subtracted.

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Problem:

Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.

Solution:

First note that it is impossible to just have the two numbers end in a 1 and a 0. This is because we have an uneven number of 1's and 0's remaining, and they can not cancel each other out to make 0. Thus, we need to add constraints to these numbers in order to achieve the task. Next note that 10_2-1_2=1_2, so an option is to force the two numbers to end in 10 and 01. This works, because now we have an even number of zeros and ones distributed among the remaining digits of the two binary integers.

Thus, when we place the integers as explained, we have 4 0s and 7 1s to distribute among the remaining 11 digits of either number. (We say either because we we will then place the digits of the other integer to match up with the digits of the first integer) Thus we can say that we are choosing 4 places out of 11 spots to put the four 1's we need to complete the requirements, and we have \displaystyle\binom{11}{4}=\boxed{330} ways to arrange the digits.

An example of two integers that work is

\setlength{\tabcolsep}{0.5mm}\begin{array}{cccccccccccccc}\\&1&0&0&1&0&0&1&0&0&1&...

Notice that 10 and 01 are the last two digits for these integers, and all the other digits match up so that they cancel when subtracted.

Posted on by djmathman | Leave a comment

2012 AIME I Problem #5

Problem: Let B be the set of all binary integers that can be written using exactly 5 zeros and 8 ones where leading zeros are allowed. If all possible subtractions are performed in which one element of B is subtracted from another, find the number of times the answer 1 is obtained.
 
First note that it is impossible to just have the two numbers end in a 1 and a 0. This is because we have an uneven number of 1's and 0's remaining, and they can not cancel each other out to make 0. Thus, we need to add constraints to these numbers in order to achieve the task. Next note that 10_2-1_2=1_2, so an option is to force the two numbers to end in 10 and 01. This works, because now we have an even number of zeros and ones distributed among the remaining digits of the two binary integers.

Thus, when we place the integers as explained, we have 4 0s and 7 1s to distribute among the remaining 11 digits of either number. (We say either because we we will then place the digits of the other integer to match up with the digits of the first integer) Thus we can say that we are choosing 4 places out of 11 spots to put the four 1's we need to complete the requirements, and we have \displaystyle\binom{11}{4}=\boxed{330} ways to arrange the digits.

An example of two integers that work is

\setlength{\tabcolsep}{0.5mm}\begin{array}{cccccccccccccc}\\&1&0&0&1&0&0&1&0&0&1&...

Notice that 10 and 01 are the last two digits for these integers, and all the other digits match up so that they cancel when subtracted.

Posted in Uncategorized | Leave a comment

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