Going through the whole Manhattan / Economist Courses and still...

Question

Dear all,

I am feeling so lost in all the topics GMAC comes up with. I am almost finished with the GMAT Economist Course and still i feel so unsure when i attempt the questions. Yesterday for example I tried the following question:

X is a three-digit positive integer in which each digit is either 1 or 2. Y has the same digits as X, but in reverse order. What is the remainder when X is divided by 3?

(1) The hundreds digit of XY is 6.
(2) The tens digit of XY is 4.

And you know what?  I had NO clue at all how to solve this  even with all the nice things everyone teaches you (POE, Ballparking and and and...). I am not a math geek - so I don't know these fancy formulas on how to solve such a question ... and the WORST thing is, I have never read about it neither in Manhattan GMAT nor in Princeton and on ECONOMIST GMAT. So the mental stress already begins with the preperation... I feel so frustrated with all that, it seems that there is no light at the end of the tunnel. Why should it be worth my while spending all my free time after work on that preperation?

Well I admit, it's kind of fun doing it, but someday i like to finish this with a score around 700. I did my first prep. test 2 weeks ago and I don't dare to share my very below average score i got. Please anyone tell me, is this for real?

My strategy is to finish this GMAT Economist tutor stuff and then I just need to do all the sample questions so I will get everything in my brain... There's just one thing, I need someone who kicks my a** ... I'm doing this alone and I feel so lost!

Thanks for any useful comments like "go for it" "work harder", "you can do it" and so on...

zero722 · Answer

Dont get discourage yet, this GMAT journey honestly is painful and there are many times anyone would want to quit. However, when that happens just relax and clear your mind, think about what really motivates you to start on this path again. Also if you get really really stuck on a question, move on, do some other question in another category and come back to it next day when you are less stressed.

Yah sometime I want to pull my hair out and just give up on all this, but then again, remember why you are here and move on...

Yah sometime I want to pull my hair out and just give up on all this, but then again, remember why you are here and move on...

jz4analytics · Answer

How I think about this questions...

Start with (2). The tens digit of XY means that both X and Y MUST have a tens digit of 2. Therefore:

X = 121, 221, 122, or 222
and 
Y = 121, 122, 221, or 222

121/3 has remainder of 1.
221/3 has remainder of 2.
122/3 has remainder of 2.
222/3 has remainder of 0.

Not sufficient. 
 
At this point you know the answer must be (A) or (C) or (E), but  not  (B) or (D).

Now if you consider (1), forgetting (2), that leaves only the pair {X=121 and Y=121} (you need the ten's digit in both to be 2 and the one's digit in both to be 1).

X can only be one number, 121, so the remainder is 1.

I think the answer is (A), (1) is sufficient.

The GMAT is made to test you on problems you haven't seen before. It's supposed to be tricky - a reasoning test - not a "can you learn how to follow a procedure" kind of math. Start thinking about these kinds of questions more analytically.

reto · Answer

Thank you. Yes indeed Stat. 1 is Sufficient. Could you explain me how you did that so easily in your head > why do i need the ten's digit to be 2 and the one's digit to be 1? What's the reason behind it? So far i got just a algebraic explanation from a certain user, which i find too time consumming on the gmat:

This rule is just based on how we multily, there is nothing more to it.

X = abc = a*100 + b*10 + c
Y = cba = c*100 + b*10 + a

XY = (a*100 + b*10 + c)*(c*100 + b*10 + a)
=>XY = 10000*ac + 1000*ab + 100a^2 + 1000bc + 100b^2 + 10ab + 100c^2 + 10*bc + ac
=> XY = 10000*ac + 1000*(ab + bc) + 100*(a^2 + b^2 + c^2) + 10*(ab + bc) + ac

If it can be proved that ac, (ab+bc) and (a^2 + b^2 + c^2) are single digits then the unit digit will be ac, 10th place will be (ab+bc) and 100th place will be (a^2 + b^2 + c^2)

thanks

Harley1980 · Answer

Hello  reto .

I can't say about other people. But my experience shows that finishing courses and reading books is like one-third of math way.

I listen Veritas course, read Manhattan books, complete Magoosh course and after all this I start doing Gmat club tests. 
This was terrible sensation. It was like completely new math. I think this is because books and courses give you the ways to solving tasks, but don't give you direct instructions about solving. 
When I have done GMAT club eighth CAT it became much easier.
 
I did 18 math cats from different sources and during last two weeks start Manhattans' tests. 
This is even more terrible than GMAT Club tests )
A lot of tasks that I completely not understand. It's really scary and unpleasant but when I reread my old drafts I understand that it's ok.
 
It's ok because you grow and GMAT give to you harder tasks, this is why it is so difficult all the time. The key moment is that you are grow. 
If you will not stop, you will beat GMAT. I am assured that this is only matter of time and your wilfulness.

jz4analytics · Answer

First off, I'm guessing this is a "hard" math question so I wouldn't worry too much right now...

Anyway, think of the multiplication. A three digit number means that you will be adding together 3 "lines" of numbers. Each line is shifted over one... and you only have to worry about the hundreds digit. Don't bother multiplying all the numbers. Since the hundreds digit is said to be 6, that will be made by adding together (1+4+1=6). The 1's and the 4 come from multiplying 1X1 and 2X2. The only way that can happen is if the 2 is in the middle, and the other digits are 1's.

Let's look at the multiplication.

121
 121 
 1 21
+2 4 20
 +12 1 00 
14641

(Hundreds digit is in bold so you can see how to visualize that place in the final answer).

Again, you do NOT have to do this out, but if you get an intuition for the numbers you can get the right answer without the work.

Note that the other combinations of multiplying digits of 1 and 2 will give you (1+1+1=3), (1+4+4=9), and (4+4+4=12)-->2 in the hundreds place. Only (1+4+1=6), and the only way to get that is (1^2)+(2^2)+(1^2), i.e. 121*121.

I did this question by thinking of all the possibilities first. I started with (Stat. 2) because it looked easier. That was easy to show it was not suff. Playing with (Stat 1) which looked more difficult eventually made it clear that only 121 for both X and Y worked, so it was sufficient.

perseverance84 · Answer

I've a question for u guys, can we consider X=112, if X=112 and Y=211 then XY=23632. Hundred's digit is 6 and the remainder when X divided by 3 is 1

jz4analytics · Answer

You're right, I forgot to check that possibility... you still end up with (1+4+1) to make the hundreds digit... but thankfully (112*211) and (121*121) have the same remainder.

Like I said, I'm sure this wasn't an easy question.

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Going through the whole Manhattan / Economist Courses and still...

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