The flowchart represents a mathematical algorithm that takes two posit

I think your second question was transcribed incorrectly--"the process" doesn't tell us WHICH process should be repeated twice!

Not that I can see. Near as I can tell, this is similar to "symbol" questions on the quantitative section--simple substitution obscured by a difficult way of expressing the math.

I understand the first value greater than zero for T is 24 as this is the first iteration but we don't output a value for T from the flow chart until b=1 so until we can get b=1 from 37 (b-1 then b/2) we never actually output a value for T. From a computer programming perspective I think this question if flawed if the answer to the first part is 24.

Hey,

Whether or not a value is actually an 'output' is irrelevant; usually, in a program, this may be implemented as showing the user a value, or perhaps passing the value to another process.

If they ask about the 'value of T', all that matters is what value it takes on in memory; if you were stepping through the program in a debugger, and you checked for what the variable 'T' contains in memory at each step, you would see all of those values changing regardless of the value actually being 'outputted'.

Hope that helps.

I agree that given the particular questions that they asked about this pattern, solving it this way is probably the easiest. If they asked questions that required looping through the block diagram many times, however, I feel that pattern recognition would be very important for maximizing time efficiency.

For example, suppose that the first question asked "What is the final value of T?" In that case, iterating through the loop would be too mechanical, and I think it would be more useful to recognize the pattern and calculate the final answer that way.

So essentially, we see that with B = 35 as an initial value, the highest power of 2 contained within 35 is 5, so we know that A will be doubled 5 times, making the final value for A equivalent to 24 * 2^5 = 24*32 = 768. From there, we recognize that A is added to T for every time B takes on an odd value: starting with 35, we can quickly see that 35 is odd, 17 is odd, and 1 is odd, so T will be composed of the first, second, and last values that are taken on by A; i.e, 24, 24*2, 24*32 = 24*35 = 840.

In fact, this algorithm just implements T = A * B, which would be crucial to recognize if we were given very large numbers. I suppose a formal proof for this is probably going too far, but here's a somewhat intuitive explanation:

Suppose we have some arbitrary value for A, and B = 61. The highest power of 2 that is less than 61 is 5 (32), so we know that A will be doubled 5 times; i.e, multiplied by 32. So to get T = A*61, we need to show that T is composed of A*32 + A*(61-32 = 29). Since we know that only A*(powers of 2) are added to T, we can break down T as follows:

61 = 32 + 16 + 8 + 4 + 1 = the sum of the following powers of 2: (0, 2, 3, 4, 5). We can see this relationship by breaking down B = 61 as follows:

61 - 1 = 60 ---------------------------------- Odd
60/2 = 30 -------------- First Halving; 2^1
30/2 = 15, 15-1 = 14 ---- 2^2 ---------------- Odd
14/2 = 7, 7-1 = 6 ------- 2^3 ---------------- Odd
6/2 = 3, 3-1 = 2 -------- 2^4 ---------------- Odd
2/2 = 1 ---------------- 2^5 ---------------- Odd

* Sorry for the poor formatting *

We can see that the number is halved 5 times, corresponding to the 5 times that A is doubled. Beside each layer is marked the corresponding 'doubling' of A.

Now, we recognize that an odd number is encountered at the 2^0, 2^2, 2^3, 2^4, and 2^5 layers, meaning we would be adding A*(1 + 4 + 8 + 16 + 32) to T, equating to 61*A.

I know that was kind of long-winded but hopefully it's useful/interesting to someone. I found it pretty neat.

　　If 35 and 27 are entered as the values for a and b, respectively, then after the process b=\(\frac{b}{2}\) is completed for the second time, the value of b is: 3, 6, 12, 13, 26.

[/quote]


Could someone help me solve the second question ?Thanks

#1 is a slightly confusing question but it is a HARD ONE. The question asks for the "First Value of T" and that assumes not actually calculating the whole process and providing output. The question is asking for Value rather than Output. This was really a "paying attention" question.

1st question, T = 24
2nd question, b = 6
As we enter the logic diagram we get T = 0, then we enter values of a and b and since b is odd, will follow T = T + a, then b = b - 1, then a = 2a and b = b/2.

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I'm still having trouble understanding why the correct answer for the second question is 3 rather than 6! Can someone please explain?

I also do not understand. I get the 6 as well;

start:
b = 27

1. b = b-1 (gives 26)
2. b/2 = 13
3. b = not 1 so repeat.
4. b still odd
5. b=b-1 (gives 12)
6. b/2= 6

So after two times this process it seems that the answer should be 6?

chetan2u Bunuel GMAT Prep shows answer as 3 for Question 2. Could you confirm?

Hi

The answer should be 6 for the question the way it is written. May be there are some changes in the wordings. Request if you can share a screenshot

Hi ya

can someone explain why the answer for the second question would be 3 rather than 6?
It is written exacly like this:

(1) If 35 and 27 are entered as the values for a and b , respectively, then after the process ​b equals b over 2​ is completed for the second time, the value of b is ()


same flowchart as posted before.

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