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A certain calculating machine has only three operation butto [#permalink]

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19 Nov 2012, 16:41

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A

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C

D

E

Difficulty:

55% (hard)

Question Stats:

63% (02:45) correct
37% (02:08) wrong based on 131 sessions

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A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A. D D D D D D R R R R R R

B. R D R D R D R D R D R D

C. A D D D D D

D. D D D D D A

E. D D D D D D D A

I have the solution and I will post if there are any inquiries, however I want to see how you guys are able to do this under 2 minutes, I could not figure it out. and I don't think the solution is that great at doing it under 2 either.

Re: A certain calculating machine has only three operation butto [#permalink]

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19 Nov 2012, 17:18

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anon1 wrote:

A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence -

D R D R D R D R D R D R ? -----choices D D D D D D R R R R R R R D R D R D R D R D R D A D D D D D D D D D D A D D D D D D D A I have the solution and I will post if there are any inquiries, however I want to see how you guys are able to do this under 2 minutes, I could not figure it out. and I don't think the solution is that great at doing it under 2 either.

While solving this under 1 miunte is possible.. typing surely isnt... I better get kudos for this

Reading question carefully is very important!

We have D, which doubles the number. A and R which add or substract 1. Now given sequence: D R D R D R D R D R D R We can take a look and know that this has to be an odd number, (Note second last D will make whatever number before that even and R will make it odd).

Now quickly go through answer choices, option B and C end with doubling operation, so these numbers will be even. Eliminate. Remaining A, D and E Look at A. After last doubling operation, there are 6 substractions. Therefore reduces even (6) from an even number (formed due to operation D)... we get even. Eliminate A. Choice D and E both end with odd numbers. so cant be eliminated in this way.

Again lets take a look at given sequence in question. There are 6 doubling operation and some substractions. Ans choice has 7 doubling operation and no substraction, therefore E would be much larger than original sequence.

We are left with D. Lets party with D because.......

Re: A certain calculating machine has only three operation butto [#permalink]

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20 Nov 2012, 15:25

Wow VIPS, brilliant work. That was very innovative insight that you used to solve the problem. The solution didn't even have that. Your ability to recognize those even/odds really was quite impressive.

Here is the solution, as you can see, is much more arduous

------official solution

To efficiently solve this question, you have to understand what the D-R operations actually do. Calculate the result of the seven times D-R, step by step:

2 -> 4 -> 3

-> 6 -> 5

-> 10 -> 9

-> 18 -> 17

-> 34 -> 33

-> 66 -> 65

The intermediate results, after each D-R, are 3,5,9,17,33,65. These numbers follow a pattern - they are all one greater than the powers of two, 2,4,8,16,32,64.

Thus, 65 is 2×2×2×2×2×2+1, i.e., five times D, followed by an A.

[[continue]]

Alternative explanation - Plug in the initial number of 2 into the answer choices, and eliminate those that do not reach the same result of 65:

(A) D D D D D D R R R R R R - the 6 Ds get you to 26 = 64, and the series of Rs reduces the result below 65.

(B) R D R D R D R D R D R D - yields a repeating pattern of 1 - 2 - 1 - 2...

(C) A D D D D D - The first A raises the initial number to 3, and the Ds ensure that the result is a multiple of 3. Since 65 is not divisible by 3, POE this answer choice.

(E) D D D D D D D A - 7 Ds is already too big - 28 = 256.

Re: A certain calculating machine has only three operation butto [#permalink]

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09 Mar 2014, 02:13

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Re: A certain calculating machine has only three operation butto [#permalink]

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03 Aug 2014, 04:12

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A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A): D D D D D D R R R R R R B): R D R D R D R D R D R D C): A D D D D D D): D D D D D A E): D D D D D D D A

Re: A certain calculating machine has only three operation butto [#permalink]

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03 Aug 2014, 08:56

1

This post was BOOKMARKED

goodyear2013 wrote:

A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A): D D D D D D R R R R R R B): R D R D R D R D R D R D C): A D D D D D D): D D D D D A E): D D D D D D D A

This is how I solved this problem:

First I tried to figure out what the original sequence result is. I didn't have to do a lot of calculating before noticing a pattern. Each R was 2x-1, so I just did every other letter. Yes, there is an algebraic way to simplify further, but in the interest of time, I didn't bother with trying to figure that out for such simple arithmetic.

From that, I got that my goal is 65. Now I had to find the correct answer choice. I start with C and move up or down depending on if I need a higher answer.

Answer choice C = (2+1)(2^5) = 96 --> too high. So I look for something that could be smaller. Option D looks smaller than C since it is powers of 2 and then adding one, instead of powers of 3.

D = 2^6 + 1

Powers of 2 should be something that is pretty easy to remember (especially if you work with computers). I know 2^6 = 64. Add 1 = 65. I have my answer: D!

For these types of questions, I find that rather than complicating it with a lot of theory, it is better to just jump into it if you are good at doing simple arithmetic in your head (or with limited scratch paper). Saves a lot of time and prevents you from making strategic errors.

A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A): D D D D D D R R R R R R B): R D R D R D R D R D R D C): A D D D D D D): D D D D D A E): D D D D D D D A

Merging similar topics. Please refer to the discussion above.
_________________

Re: A certain calculating machine has only three operation butto [#permalink]

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16 May 2016, 18:58

anon1 wrote:

A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A. D D D D D D R R R R R R

B. R D R D R D R D R D R D

C. A D D D D D

D. D D D D D A

E. D D D D D D D A

first: D4 R3 D6 R5 D10 R9 D18 R17 D34 R33 D66 R65 so we need 65 to be the answer.

A we can eliminate right away, as we have 2^6, which is way more than 65. B will yield 2. C we have 2^4, which is not sufficient. D 2^5 = 64 +1 = 65, looks good. E 2^8 - way too much

Re: A certain calculating machine has only three operation butto [#permalink]

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24 Aug 2016, 03:20

My approach was the questions demands that we find the equivalent of 65 in the alphanumeric form.

If you observe the number 65 (it is 2^6 + 1).

So it there is multiple D and an A. This can work.

So choice D has that pattern. Generally in these kind of questions, instead of focussing on a single choice, it is better to skim all choices quickly so that we pick the closest possible choice and confirm its veracity.
_________________

Aiming for a 3 digit number with 7 as hundredths Digit

A certain calculating machine has only three operation buttons, D, R and A. The D operation doubles the number the machine displays, the R operation subtracts 1 from the number the machine displays, and the A operation adds 1 to the number the machine displays. If the machine initially displays the number 2, which of the following operation sequences (read from left to right) would lead to the same final displayed result as the sequence - D R D R D R D R D R D R ?

A. D D D D D D R R R R R R

B. R D R D R D R D R D R D

C. A D D D D D

D. D D D D D A

E. D D D D D D D A

This is definitely possible in under 2 minutes, but you have to spend some time understanding it before launching into the math. It's a good example of a problem that rewards you for slowing down at the beginning.

Take a look at the original sequence. You're alternating between doubling the number, and subtracting 1 from it. Interestingly, that means you're bouncing back and forth between even and odd values (when you double a value it becomes even, and when you subtract 1, it becomes odd.) So, the sequence will end on an odd number.

That immediately eliminates B and C, which always end on an even number. With a little more thought, you can eliminate A: after the last doubling operation, which leaves you with an even number, you subtract 1 six times. So, the result will also be even.

You're down to just D and E. The only difference between these is the number of times the value is doubled. In the original sequence, we double the value six times. However, between each doubling operation, we subtract 1. That means that the original sequence will definitely give an answer that's smaller than simply doubling the number 6 times. Answer choice E, on the other hand, gives a value that's larger than doubling the number 6 times. The only remaining possibility is D.

At this point, I'd base my answer on how much time I had left. If I was tight on time, I'd pick D and move on. If not, I'd work out the math first to double check my logic:

using the original sequence: D R D R D R D R D R D R = 2-4-3-6-5-10-9-18-17-34-33-66-65 using the sequence in D: D D D D D A = 2-4-8-16-32-64-65

Looks right!
_________________

Chelsey Cooley | Manhattan Prep Instructor | Seattle and Online

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