CP #24

MGMAT CAT#2

Is the positive integer N a perfect square?

(1) The number of distinct factors of N is even.
(2) The sum of all distinct factors of N is even.

Answer Explanation:
(1) SUFFICIENT: The factors of any number N can be sorted into pairs that multiply to give N. (For instance, the factors of 24 can be paired as follows: 1 and 24; 2 and 12; 3 and 8; 4 and 6.) However, if N is a perfect square, one of these ‘pairs’ will consist of just one number: the square root of N. (For example, if N were 49, it would ahve the factor pair 7 × 7) Since all of the other factors can be paired off, it follows that if N is a perfect square, then N has an odd number of factors. (If N is not a perfect square, then all of its factors can be paired off, so it will have an even number of factors.) This statement then implies that N is not a perfect square.

(2) SUFFICIENT: Let N be a perfect square. If N is odd, then all factors of N are odd. Therefore, by the above reasoning, N has an odd number of odd factors, so their sum must be odd. If N is even, let M be the product of all the odd prime factors (as many times as they appear in N – not distinct) of N, which is also a perfect square. (For instance, if N = 100, then M =5 × 5 = 25.) Then the sum of factors of M is odd, by the above reasoning. Furthermore, all other factors of N (i.e., that don’t also divide M) are even. The sum total is thus odd + even = odd.
Therefore, the sum of the factors of any perfect square is odd, so this statement implies that N is not a perfect square.
This statement can also be investigated by trying several cases of even perfect squares (4, 16, 36, 64, 100), noting that in each case the sums of the factors are odd, and generalizing.

The correct answer is D.

-----------

I picked numbers to get to my answer, but made a silly mistake by omitting "1" as a distinct factor in Statement II.

And to the MGMAT explanation - hunh?!

OG12 quant almost done

40 more questions from OG12 done. 19/20 right in PS (avg time: 1:56) and 18/20 right in DS (avg time: 2:16).

I feel more focus was in order while doing DS questions, but there are probably more concepts I need to understand in greater depth in order to get my questions right at a quicker pace.

GMAT Love!

Back to OG12 Quant

Wohkay! Seems like I've been out of touch with OG type Math questions for too long.
Did 20 PS and 20 DS questions (timed) today.
Got 4 wrong in PS and...wait for it...9 wrong in DS! Eeeks!
Gotta review and get back to official questions! Good thing I have only the medium-difficult questions remaining... :)

GMAT love to the rest of you! Hope you're studying hard for the next 72 hours...to be able to party tomorrow night without guilt! :)

CP#17

Question (from MGMAT Practice Test #1)
For positive integer k, is the expression (k + 2)(k² + 4k + 3) divisible by 4?

(1) k is divisible by 8.

(2)

k + 1 3

is an odd integer.

Solution
The quadratic expression k² + 4k + 3 can be factored to yield (k + 1)(k + 3). Thus, the expression in the question stem can be restated as (k + 1)(k + 2)(k + 3), or the product of three consecutive integers. This product will be divisible by 4 if one of two conditions are met:

If k is odd, both k + 1 and k + 3 must be even, and the product (k + 1)(k + 2)(k + 3) would be divisible by 2 twice. Therefore, if k is odd, our product must be divisible by 4.

If k is even, both k + 1 and k + 3 must be odd, and the product (k + 1)(k + 2)(k + 3) would be divisible by 4 only if k + 2, the only even integer among the three, were itself divisible by 4.

The question might therefore be rephrased “Is k odd, OR is k + 2 divisible by 4?” Note that a ‘yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a ‘no’ to both conditions.

(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive ‘no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.

(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.

The correct answer is A.


Query
I think the answer is D. In Statement (2), for (k+1)/3 to be an odd INTEGER, k+1 has to be divisible by 3. Also, k must be even. Therefore, k could be 2,8,... And if k is an even integer, (k + 1)(k + 2)(k + 3) will not be divisible by 4 for any of these values of k. Therefore, Statement (2) also gives us the answer. Please tell me where I'm going wrong.

Disturbed.

RC accuracy remains intact with 6/6 answers right.

PS - need to review Number Properties (esp Remainders).

DS - accuracy fell massively - 4/10 wrong.

CR - got 2/10 wrong.

Overall, getting better. Feel better. But, need to review old concepts. AND not be disturbed while doing my questions!!!

Also, got access to BTG account as a winner of BTG 100k Challenge! :)

Will also write a review on new & improved BTG premium questions by end of this week!

Stay tuned fellow gmatters!

CP #10

GMAT Blue Quiz - Kaplan

UNDERSTAND EXPLANATION!!!

Question

Is a – b > 0?

(1)    a > 2b

(2)    b > a + 3

Solution

The inequality in the question stem can be manipulated so that the question reads "is a > b?"

At first glance, Statement (1) would seem to be sufficient. If a and b are both positive, and then a is certainly greater than b. But what if a and b are negative? If a = -3 and b = -2? Since -3 > -4, a would be greater than 2b; but since -2 > -3, b would be greater than a. Since Statement (1) can lead to either a yes or no answer, it is insufficient.

Manipulating the inequality in Statement (2) is more fruitful. Subtracting b and 3 from both sides gives you -3 > a-b. If that's true, then a – b is certainly less than zero. Statement (2) leads to a single, definite answer (which happens to be "no") and so is sufficient.

Kaplan Permier: Blue Quiz

Completed Kaplan Premier's Blue Quiz today. Scored an 87% overall. Got 26/30 right. Wrong - 2 CR (DOUBT IN #20 - ASK A KAPLAN INSTRUCTOR), 1 SC (idiom - as many...as) and 1 DS (ASK FRIEND!).

Kaplan Diagnostic Quiz & Speed Math

Hi...

Work has kept me quite busy, so had left GMAT studies for sometime. It's time now.

Recap

Took the GMAT Diagnositc Quiz on Kaplan Premier's online companion. Scored 89% - 34 / 38 were correct answers. Got 1 wrong each in DS, PS, RC and CR. It was untimed.

To Do

While reviewing, there is 1 DS question I could not understand - will ask a friend - also, need to put it as a CP.

Another DS question took too long - must review that.

RC Strategy

In RC, writing what the para is talking about as short notes helps me. It's a PR method, but it seems to be working. It doesn't work in those passages where I have literally been unable to comprehend the meaning. To reduce the probability of that happening, I need to start taking out time to read (I used to be a voracious reader till time constrains brought my reading time down to near negligible). I also need to work on my vocab - I have Word Power Made Easy (WPME) by Norman Lewis. Maybe I should revisit it.

Speed Math

Have also been learning calculation shortcuts from Bill Handley's Speed Mathematics. It's actually quite fun and I've done a few chapters.

To Do

Do the chapter on decimal multiplication. Practise calculations.

A silly mistake (Quant)

From vaibhav.iit2002 on his BTG thread: 51 in QA - debrief.

"In DS, problems always think TWICE before marking C or E. Its a common GMAT problem to give an impression that you need both A and B to solve the problem while you can solve just by using A or B[both integers].
e.g. 2a + 3b = 10
21a + 23b = 67[i don't exactly remember the example]
Its very very easy to mark C as we think that we need 2 equations for 2 unknown variables. Try putting values for a and b[if feasible] in equation 2 as only a=1 and b=2 will fit."

CP #2

Wicked question by MGMAT this week! Check it out...


MGMAT Challenge Problem of the Week

10/19/09

Question

If a, b, and c are positive integers, with a < b < c, are a, b, and c consecutive integers?

(1) 1/a – 1/b = 1/c

(2) a + c = b² – 1

(A)	Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B)	Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C)	BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D)	EACH statement ALONE is sufficient.
(E)	Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.

Answer

The question can be rephrased "Is b = a + 1 and is c = a + 2?"

One way to approach the statements is to substitute these expressions involving a and solve for a. Since this could involve a lot of algebra at the start, we can just substitute a + 1 for b and test whether c = a + 2, given that both are integers.

Statement 1: SUFFICIENT.
Following the latter method, we have
1/a – 1/(a + 1) = 1/c
(a + 1)/[a(a + 1)] – a/[a(a + 1)] = 1/c
1//[a(a + 1)] = 1/c
a² + a = c

Now we substitute a + 2 for c and examine the results:
a² + a = a + 2
a² = 2
a is the square root of 2. However, since a is supposed to be an integer, we know that our assumptions were false, and a, b, and c cannot be consecutive integers.

We can now answer the question with a definitive "No," making this statement sufficient.

We could also test numbers. Making a and b consecutive positive integers, we can solve the original equation (1/a – 1/b = 1/c). The first 4 possibilities are as follows:
1/1 – 1/2 = 1/2
1/2 – 1/3 = 1/6
1/3 – 1/4 = 1/12
1/4 – 1/5 = 1/20

Examining the denominators, we can see that c = ab. None of these triples so far are consecutive, and as a and b get larger, c will become more and more distant, leading us to conclude that a, b, and c are not consecutive.

Statement 2: SUFFICIENT
Let's try substituting (a + 1) for b and (a + 2) for c.

a + a + 2 = (a + 1)² – 1
2a + 2 = a² + 2a
2 = a²

Again, we get that a must be the square root of 2. However, we know that a is an integer, so the assumptions must be false. We can answer the question with a definitive "No," and so the statement is sufficient.

The answer is D: Each statement is sufficient.
--------------------------------------------------------------
The first option required just a little more calculation that the second.
I did get the answer on my first try, but I still wanted to post it here because the logic that struck me while trying to solve it may not have struck me at another point of time. I know I've opened my mind to newer ways of solving questions. Earlier, even if the idea would've struck me, just knowing it was a little bit more complicated than what I'm extremely comfortable with, I would've confused myself. Now, I take a deep breath, try to understand why I'm thinking that way, and the answer comes. Even if it doesn't, atleast I'm aware of the logic I used! I should be thanking the people at T.I.M.E classes for this. The Math teacher in Delhi (Kailash Colony) as well as Bombay (Charni Road) were pretty damn good. And they have helped me think of Math in a simpler, clearer manner, and I think I'm getting over my fear now, and beginning to not underestimate myself.
I think I've left my notepad at home. I should write my thought process here.

Challenge Problem (CP) #1

From now on, I'll post the problems I find on the net that I find challenging, didn't get in the first try or didn't solve by the better method. Here's the first one!

NOTE: Whenever the problem is from MGMAT's CP of the Week, click on the link to go the original source of the problem.

MGMAT Challenge Problem of the Week


10/05/09

Question

Xander, Yolanda, and Zelda each have at least one hat. Zelda has more hats than Yolanda, who has more than Xander. Together, the total number of hats the three people have is 12. How many hats does Yolanda have?

(1) Zelda has no more than 5 hats more than Xander.

(2) The product of the numbers of hats that Xander, Yolanda, and Zelda have is less than 36.

Answer

Write x for the number of hats Xander has, y for the number of hats Yolanda has, and z for the number of hats Zelda has. From the question stem, we know that x < y < z and that x + y + z = 12. Moreover, since each person has at least one hat, and people can only have integer numbers of hats, we know that x, y, and z are all positive integers. With this number of constraints, we should go ahead and list scenarios that fit all the constraints. Start with x and y as low as possible, then adjust from there, keeping the order, keeping the sum at 12, and ensuring that no two integers are the same.

Scenario	x	y	z
(a)	1	2	9
(b)	1	3	8
(c)	1	4	7
(d)	1	5	6
(e)	2	3	7
(f)	2	4	6
(g)	3	4	5

These are the only seven scenarios that work. As a reminder, we are looking for the value of y. Now, we turn to the statements.

Statement (1): INSUFFICIENT. We are told that z – x is less than or equal to 5. This rules out scenarios (a) through (c), but the last four scenarios still work. Thus, y could be 3, 4, or 5.

Statement (2): INSUFFICIENT. We are told that xyz is less than 36. We work out this product for the seven scenarios:
(a) 18
(b) 24
(c) 28
(d) 30
(e) 42
(f) 48
(g) 60

We can rule out scenarios (e) through (g), but (a) through (d) still work. Thus, y could be 2, 3, 4, or 5.

Statements (1) and (2) together: SUFFICIENT. Only scenario (d) survives the constraints of the two statements. Thus, we know that y is 5.

The correct answer is (C): BOTH statements TOGETHER are sufficient to answer the question, but neither statement alone is sufficient.