## Am I solving these correctly?

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Am I solving these correctly?

I would appreciate any feedback concerning if I am solving these problems in an optimal manner. Based on the feedback I have been getting on other problems, I wouldn't be surprised if there is a better way to solve the following problems than the way I am solving them. Thanks.

Problem 1) (this problem is from practice test 1 on the Computer Adaptive General GRE)
"The 'reflection' of a positive integer is obtained by reversing its digits. For example, 321 is the reflection of 123. The difference between a five digit integer and its reflection must be divisible by which number?"

1) 2
2) 4
3) 5
4) 6
5) 9

I correctly answered this as choice (5). My strategy for solving this was to come up with five digit numbers, where the difference between the number and it's reflection was not divisible by choices (1) through (4). By process of elimination the answer had to be choice (5). My concern is, however, that there may be a faster way, that if I tried my method on the actual test, I would not have enough time to finish. Is there a better way to solve this?
(continued in post 2 below)
Last edited by cooper on Mon Aug 24, 2009 8:21 pm, edited 1 time in total.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

A second, more serious, problem I had (also from practice test 1) is:

a>0 and b>0

col. A: [a + b + 2(ab)^(1/2)]^(1/2)
col. B: a^(1/2) + b^(1/2)

The answer is "C", both values are equal.

My strategy for solving this was to insert values into a and b, both positive integers and positive fractions. I kept getting the same answer so I guessed that the correct answer is "C" (which it is). I say I guessed because I couldn't swear that every choice for a and b would give the same amount for both columns just because several times I did get the same answer. Can anyone think of a better way to solve this? Thanks again.

sravanskarri
Posts: 58
Joined: Sat Jun 14, 2008 10:19 pm

### Re: Am I solving these correctly?

This pretty straight forward ... expansion of (x+y ) ^ 2 where x = sqrt(a), y = sqrt(b).. I hope you are not kidding . .

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

sravanskarri wrote:This pretty straight forward ... expansion of (x+y ) ^ 2 where x = sqrt(a), y = sqrt(b).. I hope you are not kidding . .

Thanks for the help sravanskarri, I simply did not see that relationship (despite the fact that I have been looking at that problem several times over the past several days).

If you don't mind, can you give me feedback on the first problem?

Posts: 41
Joined: Sat Dec 06, 2008 11:10 am

### Re: Am I solving these correctly?

cooper wrote:I would appreciate any feedback concerning if I am solving these problems in an optimal manner. Based on the feedback I have been getting on other problems, I wouldn't be surprised if there is a better way to solve the following problems than the way I am solving them. Thanks.

Problem 1) (this problem is from practice test 1 on the Computer Adaptive General GRE)
"The 'reflection' of a positive integer is obtained by reversing its digits. For example, 321 is the reflection of 123. The difference between a five digit integer and its reflection must be divisible by which number?"

1) 2
2) 4
3) 5
4) 6
5) 9

I correctly answered this as choice (5). My strategy for solving this was to come up with five digit numbers, where the difference between the number and it's reflection was not divisible by choices (1) through (4). By process of elimination the answer had to be choice (5). My concern is, however, that there may be a faster way, that if I tried my method on the actual test, I would not have enough time to finish. Is there a better way to solve this?
(continued in post 2 below)

I just worked this out in roughly 15 seconds. I think it is pretty neat. Any finite N-digit number can be written as

q = a_0 10^0 + ... + a_k 10^k + ... a_(N-1) 10^(N-1)

where a_i = 1,2,...,9. hence the reflection would be written

q' = a_0 10^(N-1) + ... + a_k 10^(N-1-k) + ... a_(N-1) 10^0.

Noting (10 mod 9 = 1), we have (10^k mod 9 = 1^k = 1), we compute

(q - q') mod 9 = a_0 + ... + a_(N-1) - (a_0 + ... + a_(N-1)) = 0

meaning 9 always divides q - q'. This works for all length numbers, not just 5 digit numbers. This works by the same mechanism that (the sum of the digits of a number q divides 3) --> (q is divisible by 3), interestingly enough.

twistor
Posts: 1531
Joined: Thu Apr 13, 2006 2:47 pm

### Re: Am I solving these correctly?

I'm not sure what WakkaDojo is claiming.... he seems to be saying all numbers are divisible by 9. Anyway, you can safely ignore his solution since you will not be required to know modulo math for the general GRE.

There is a much simpler, nicer solution that is very direct. All it requires is that you know how to represent a 5 digit number in the decimal system and that you know how to subtract two numbers.

First, any five digit number is represented as;

a_0*10^0 + a_1*10^1 + a_2*10^2 + a_3*10^3 + a_4*10^4

It's reflection, therefore, is:

a_4*10^0 + a_3*10^1 + a_2*10^2 + a_1*10^3 + a_0*10^4

Now, subtracting them you get:

(a_0 - a_4)*10^0 + (a_1 - a_3)*10^1 + (a_3 - a_1)*10^3 + (a_4 - a_0)*10^4

Note the symmetry here: a_1 - a_3 = - (a_3 - a_1) and a_0 - a_4 = - (a_4 - a_0).
With that in mind, we write:

-(a_4 - a_0)*10^0 - (a_3 - a_1)*10^1 + (a_3 - a_1)*10^3 - (a_4 - a_0)*10^4

= (a_4 - a_0)(10^4 - 10^0) + (a_3 - a_1)*(10^3 - 10^1)

= (a_4 - a_0)(9999) + (a_3 - a^1)(990) = 9*[(a_4 - a_0)(1111) + (a_3 - a_1)(110) ]

Thus the result is divisible by 9.
Last edited by twistor on Tue Aug 25, 2009 10:08 am, edited 1 time in total.

nathan12343
Posts: 249
Joined: Sat Dec 20, 2008 9:34 am

### Re: Am I solving these correctly?

I wonder which is faster, dividing some random 5-digit number and its reflection by each of the possible solutions or coming up with the formal solution like Twistor. You should be able to do long division on five digit numbers quickly.

I doubt WakkaDojo's method would be very useful on the test since you would need to do this manipulation for each answer.

*Edited for stupidity.
Last edited by nathan12343 on Tue Aug 25, 2009 2:01 am, edited 1 time in total.

Posts: 41
Joined: Sat Dec 06, 2008 11:10 am

### Re: Am I solving these correctly?

twistor wrote:I'm not sure what WakkaDojo is claiming.... he seems to be saying all numbers are divisible by 9.

Absolutely not. Read closer, the catch is that the factors of 10^k all become 1 and the coefficients a_k cancel.

grae313
Posts: 2297
Joined: Tue May 29, 2007 8:46 pm

### Re: Am I solving these correctly?

Also, you can eliminate options 1), 2), and 3) right off the bat since you can very quickly think of exceptions. Between 6 and 9, 9 is just the obvious choice because there are always funny little "9" tricks in a base 10 counting system Done! (sort of) Actually, cooper, if I had that question during a test I'd eliminate the first three immediately then try a 5 digit number like you did to make sure.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

nathan12343 wrote:I wonder which is faster, dividing some random 5-digit number and its reflection by each of the possible solutions or coming up with the formal solution like Twistor. You should be able to divide five digit numbers quite quickly, especially since you have a calculator.

I doubt WakkaDojo's method would be very useful on the test since you would need to do this manipulation for each answer.

You're not allowed to use a calculator on the exam. However, it's not too difficult. What I did (admittedly I did this after the test was over when I thought carefully about it, I don't know if I could think of this during an actual exam) was to think of a number, like 54321, then it's reflection, 12345. I subtracted, got 41976 and realized that choice (3) was eliminated. Then I decided to substitute a 6 for a 5 (so now the number is 64321) and I quickly saw that the solution would have a 5 in the units place and this clearly eliminates choices (1), (2) and (4). This left me with choice (5). It also left me with a cold feeling that this approach might not work on the exam if the question was given to me slightly differently and that there is a better solution that I don't know about (but the students who get 800 on the exam do know about).
Last edited by cooper on Tue Aug 25, 2009 12:38 am, edited 1 time in total.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

grae313 wrote:Also, you can eliminate options 1), 2), and 3) right off the bat since you can very quickly think of exceptions. Between 6 and 9, 9 is just the obvious choice because there are always funny little "9" tricks in a base 10 counting system Done! (sort of) Actually, cooper, if I had that question during a test I'd eliminate the first three immediately then try a 5 digit number like you did to make sure.

You may very well have had that question Jessie. One day before you took the exam you gave yourself two computer tests, and scored 800 on both. I presume you took the same test I did (I guess it could have been changed in the last two years or so). Maybe the computer didn't give you that question (or maybe you forgot if it did), but you might have been given that question.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

twistor wrote:I'm not sure what WakkaDojo is claiming.... he seems to be saying all numbers are divisible by 9. Anyway, you can safely ignore his solution since you will not be required to know modulo math for the general GRE.

There is a much simpler, nicer solution that is very direct. All it requires is that you know how to represent a 5 digit number in the decimal system and that you know how to subtract two numbers.

First, any five digit number is represented as;

a_0*10^0 + a_1*10^1 + a_2*10^2 + a_3*10^3 + a_4*10^4

It's reflection, therefore, is:

a_4*10^0 + a_3*10^1 + a_2*10^2 + a_1*10^3 + a_0*10^4

Now, subtracting them you get:

(a_0 - a_4)*10^0 + (a_1 - a_3)*10^1 + (a_3 - a_1)*10^3 + (a_4 - a_0)*10^4

Note the symmetry here: a_1 - a_3 = - (a_3 - a_1) and a_0 - a_4 = - (a_4 - a_0).
With that in mind, we write:

-(a_4 - a_0)*10^0 - (a_3 - a_1)*10^1 + (a_3 - a_1)*10^3 - (a_4 - a_0)*10^4

= (a_4 - a_0)(10^4 - 10^0) + (a_3 - a_1)*(10^3 - 10^1)

= (a_4 - a_0)(9999) + (a_3 - a^1)(990) = 9*[(a_4 - a_0)(1111) + (a_3 - a_1)(99) ]

Thus the result is divisible by 9.

I follow your reasoning twistor (although you did make two or so typos. For example, in the final answer, the "99" should be a "110"), but did you really think like this when you took the practice exam (or, if you did take the exam and were given this question, do you think you would have thought this solution up)? During the one or two minutes you have to answer the question, did you really come up with this complicated solution? It seems you would have to be a genius to think of something like this. It reminds me of that cute story about the boy who very quickly completed his punshment of adding all numbers between 1 and 100. He did it quickly because he thought of a short cut. However, he went on to become a famous mathematician, and his short cut is now a proof. I am sure that coming up with a solution like that doesn't happen every day.

twistor
Posts: 1531
Joined: Thu Apr 13, 2006 2:47 pm

### Re: Am I solving these correctly?

No.

Sorry about the typos. It was late and I was in a hurry.

Though I do work problems out like this given enough time, I was simply far too nervous on the actual exam to carry out any kind of formal procedure.

I probably would've done trial and error, and after two or three would've guessed the answer (probably wrong) and moved on.

Which is why I think the general (and subject) GRE is bullshit. I'm quite clearly capable of providing a formal proof of the fact but on the test I'd likely have gotten it wrong due to anxiety and time constraints.

Of course, it only took me about 2 minutes to do that solution. That's probably too much time for the general test, but I can easily see a "guess-and-check" type method taking just as long. The thing is that you really can't practice for these, no matter how much you practice. That's because the problems they throw at you will require you to come up with solutions on the spot -- your pre-practiced solutions just won't work.

I guess checking two or three numbers would be the fastest method. 2, 4 and 6 are very easy to eliminate if your answers aren't even. Only numbers that end in 5's or 0's can be divisible by 5. So that *might* leave you with 9 if you guess the right numbers.

Another way would be to apply a divisibility test:

http://www.jimloy.com/number/divis.htm

grae313
Posts: 2297
Joined: Tue May 29, 2007 8:46 pm

### Re: Am I solving these correctly?

twistor wrote:I guess checking two or three numbers would be the fastest method. 2, 4 and 6 are very easy to eliminate if your answers aren't even. Only numbers that end in 5's or 0's can be divisible by 5. So that *might* leave you with 9 if you guess the right numbers.

Oh, duh, of course six is easy to eliminate too! So any number that begins with an odd number, you know that it's reflection can't be divisible by 2, 4, or 6, and your number would have to begin in 5 for it to be divisible by 5, leaving 9 as the only possible choice.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

twistor wrote:No.

Sorry about the typos. It was late and I was in a hurry.

Though I do work problems out like this given enough time, I was simply far too nervous on the actual exam to carry out any kind of formal procedure...

You would have guessed and gotten it wrong? But twistor, didn't you get an 800 on the Quantitative (like just about everyone else but me)? Maybe not.

Anyway, the GRE is probably largely useless like you said. In 1995 I obtained a very high score on the Verbal GRE, and I was therefore admitted into a PhD program in Psychology with a Fellowship. Everyone, particularly my girlfriend, was very proud of me (I will never forget the proud look on my girlfriend's face when I told her, knowing that she was very proud of me was quite simply the most wonderful feeling I ever had ), especially when I told them that the GRE correlates highly with IQ tests. The joke was that all I really did was memorize Barron's 3500 vocabulary word list, and that's why I did so well on the exam. I personally don't see why being a walking lexicon (lexicon means dictionary, I learned that from the word list ) makes you more capable of studying Psychology.

By the way, all I did was apply the divisibility test, and I was able to reject all the other choices that way (6 was a little harder, but I was able to reject that one too with this method just like grae is saying).

twistor
Posts: 1531
Joined: Thu Apr 13, 2006 2:47 pm

### Re: Am I solving these correctly?

No, I got a far more realistic 700-something.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

twistor wrote:No, I got a far more realistic 700-something.

If you don't mind my asking, what exactly did you get? If it was below 750, how hard was it to get accepted into a program (assuming you were accepted)? Do you think I can get into a PhD program with a 750? I don't mind not getting a Fellowship again, but I would like to know I can get in (assuming I decide to apply, I still haven't decided if it's practical for me to do this). Thanks for any feedback you're willing to give.

twistor
Posts: 1531
Joined: Thu Apr 13, 2006 2:47 pm

### Re: Am I solving these correctly?

I took it twice. The first time it was something like 690 and the second time I think it was 710. I applied to the two best medical physics programs in the country and was accepted to both. To be fair, I was highly involved in research and had excellent letters of recommendation and nearly a perfect GPA. I think that if the rest of your application is outstanding it's very easy to overlook minor things like your general GRE scores. If the rest of you're application says your outstanding and your general GRE score is the one anomalous blemish on your record then I wouldn't worry too much about it.

However, if your application shows you are a mediocre candidate, and your scores say the same, then that significantly reduces your chances.

Don't worry about not getting into a program with a 750. 750 is still a very good score. You will almost certainly be accepted somewhere... it may not be Harvard, but it will be somewhere.

cooper
Posts: 59
Joined: Wed Jul 08, 2009 4:57 am

### Re: Am I solving these correctly?

twistor wrote:I took it twice. The first time it was something like 690 and the second time I think it was 710...

Okay twistor, thanks for the advice. As I said, it remains to be seen if I decide to apply. I wish I were in my 20's again (I guess all people in their 40's wish that), then I would get my PhD in Physics and spend, hopefully, the next 40+ years conducting research in Quantum Mechanics or one of the other exciting areas of Physics. But at 43, given my life circumstances, when I recently completed my Bachelors in Physics and successfully finished an Engineering Internship, I decided, for practical reasons, that it would be better to Plan B it, and work as an Engineer. When the recession made entry level Engineering positions scarce, I thought seriously about going for a Masters in Physics while I am waiting for the economy to improve. Not a bad plan, but I am still not sure it's practical given my life situation. Oh well, I guess I will wait and see how things work out. Thanks again.