sravanskarri wrote:This pretty straight forward ... expansion of (x+y ) ^ 2 where x = sqrt(a), y = sqrt(b).. I hope you are not kidding . .
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?"
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)
twistor wrote:I'm not sure what WakkaDojo is claiming.... he seems to be saying all numbers are divisible by 9.
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.
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.
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.
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.
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...
twistor wrote:No, I got a far more realistic 700-something.
twistor wrote:I took it twice. The first time it was something like 690 and the second time I think it was 710...
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