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11 Nov 2022, 08:22
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Difficulty:
35%
(medium)
Question Stats:
73% (01:56) correct
27%
(02:40)
wrong
based on 56
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If 1 < a < b < c < d, what is the value of d?
(1) a + b + c + d = 170
(2) a, b, c, and d are all positive integers in the set of numbers that can be written in the form of 2^n, where n is also a positive integer.
(1) a + b + c + d = 170
(2) a, b, c, and d are all positive integers in the set of numbers that can be written in the form of 2^n, where n is also a positive integer.
11 Nov 2022, 22:15
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IMO a good question ! The question would have been a bit more challenging if this had been a PS question 
What we know
The values when plotted over a number line look as follows
--------0--------1--------a--------b--------c--------d--------
Statement 1
We know that the sum of a + b + c + d = 170
However, we do not know the individual positions. For example, d can take any value and we can adjust the value of a , b and c accordingly to meet the criteria.
Hence this statement is not sufficient.
Statement 2
a, b, c, and d are all positive integers in the set of numbers that can be written in the form of 2^n, where n is also a positive integer.
Just as in statement 1, the details are not sufficient to find the value of d. From the question stem, we know the relative position of a, b , c and d with respect to each other. However, we can have multiple such values based on the value of power of 2.
Combined
The statements when combined get interesting. We know that the sum of the numbers is 170 and the numbers are powers of 2.
So let's write the numbers as powers of 2 and see if we can get some meaning out of it -
\(a = 2^x; b=2^y; c=2^z; d=2^p\)
\(2^{x} + 2^{y} + 2^{z} + 2^{p} = 170\)
Because the numbers are all even, the sum must also be even.
170 can be represented as 2 * 85
If take 2 common the LHS
\(2* (2^{x-1} + 2^{y-1} + 2^{z-1} + 2^{p-1}) = 2 * 85\)
Hence
\(2^{x-1} + 2^{y-1} + 2^{z-1} + 2^{p-1} = 85\)
Now the sum of powers of two is, hence one of the numbers must be odd and it needs to be 1. The only possibility is of that to occur is when \(2^{x-1}\) = 1, as it is the smallest number (remember the relative order from statement 1)
\(2^{x-1}\) = \(2^0\)
So \(x-1\) = 0
\(x = 1\)
So we are able to the value of the powers. One can continue this process* and get the value of d (as its a DS question, we need not solve to get the value of d but we have all the information required if need be). The combination is sufficient.
IMO C
*I have added the rest of the process under spoiler.
What we know
The values when plotted over a number line look as follows
--------0--------1--------a--------b--------c--------d--------
Statement 1
We know that the sum of a + b + c + d = 170
However, we do not know the individual positions. For example, d can take any value and we can adjust the value of a , b and c accordingly to meet the criteria.
Hence this statement is not sufficient.
Statement 2
a, b, c, and d are all positive integers in the set of numbers that can be written in the form of 2^n, where n is also a positive integer.
Just as in statement 1, the details are not sufficient to find the value of d. From the question stem, we know the relative position of a, b , c and d with respect to each other. However, we can have multiple such values based on the value of power of 2.
Combined
The statements when combined get interesting. We know that the sum of the numbers is 170 and the numbers are powers of 2.
So let's write the numbers as powers of 2 and see if we can get some meaning out of it -
\(a = 2^x; b=2^y; c=2^z; d=2^p\)
\(2^{x} + 2^{y} + 2^{z} + 2^{p} = 170\)
Because the numbers are all even, the sum must also be even.
170 can be represented as 2 * 85
If take 2 common the LHS
\(2* (2^{x-1} + 2^{y-1} + 2^{z-1} + 2^{p-1}) = 2 * 85\)
Hence
\(2^{x-1} + 2^{y-1} + 2^{z-1} + 2^{p-1} = 85\)
Now the sum of powers of two is, hence one of the numbers must be odd and it needs to be 1. The only possibility is of that to occur is when \(2^{x-1}\) = 1, as it is the smallest number (remember the relative order from statement 1)
\(2^{x-1}\) = \(2^0\)
So \(x-1\) = 0
\(x = 1\)
So we are able to the value of the powers. One can continue this process* and get the value of d (as its a DS question, we need not solve to get the value of d but we have all the information required if need be). The combination is sufficient.
IMO C
*I have added the rest of the process under spoiler.
Curious .. eh ! 
\(1 + 2^{y-1} + 2^{z-1} + 2^{p-1} = 85\)
\(2^{y-1} + 2^{z-1} + 2^{p-1} = 84\)
\(2^2*(2^{y-3} + 2^{z-3} + 2^{p-3}) = 4*21\)
\(2^2*(2^{y-3} + 2^{z-3} + 2^{p-3}) = 4*21\)
\(2^{y-3} + 2^{z-3} + 2^{p-3} = 21\)
Again we see that the sum of some powers of 2 is odd. So \(2^{y-3} = 1\)
\(1 + 2^{z-3} + 2^{p-3} = 21\)
\(2^{z-3} + 2^{p-3} = 20\)
\(2^2 * (2^{z-5} + 2^{p-5})= 4 * 5 \)
\(2^{z-5} + 2^{p-5} = 5 \)
\(2^{z-5}\) =1 (following the same logic)
\(1 + 2^{p-5} = 5 \)
\( 2^{p-5} = 4 \)
p-5 = 2
p = 7
\(d = 2^p = 2^7 = 128\)
\(1 + 2^{y-1} + 2^{z-1} + 2^{p-1} = 85\)
\(2^{y-1} + 2^{z-1} + 2^{p-1} = 84\)
\(2^2*(2^{y-3} + 2^{z-3} + 2^{p-3}) = 4*21\)
\(2^2*(2^{y-3} + 2^{z-3} + 2^{p-3}) = 4*21\)
\(2^{y-3} + 2^{z-3} + 2^{p-3} = 21\)
Again we see that the sum of some powers of 2 is odd. So \(2^{y-3} = 1\)
\(1 + 2^{z-3} + 2^{p-3} = 21\)
\(2^{z-3} + 2^{p-3} = 20\)
\(2^2 * (2^{z-5} + 2^{p-5})= 4 * 5 \)
\(2^{z-5} + 2^{p-5} = 5 \)
\(2^{z-5}\) =1 (following the same logic)
\(1 + 2^{p-5} = 5 \)
\( 2^{p-5} = 4 \)
p-5 = 2
p = 7
\(d = 2^p = 2^7 = 128\)
