MANHATTAN GMAT PROBLEM OF THE WEEK

Sometimes, I wonder about the advertising of these sites... this is a serious question in number theory and it could be put in an induction proof on a Discrete structures final.

Well, after looking at the first few numbers in the sequence

t(0) = 3
t(1) = 3 + 1
t(2) = 3 + 1 + 2
t(3) = 3 + 1 + 2 + 3
t(4) = 3 + 1 + 2 + 3 + 4

We can summarize that each term is 3 + (n+1)n/2 which I will call (*).

So we now look at the questions

Statement A tells you n+1 is divisible by 3 - now if, (n+1)*n/2 can be even, then A is insufficient. And I think you can do that by making n+1 divisible by both 4 and 3 (for example, 12).

Statement B is a bit tougher - I did it by rewriting (*) as 3 +n(n-1)/2 + n

Now, if n-1 is divisible by 4, then n is odd. and if n-1 is divisible by 4, n(n-1)/2 is even, so B is sufficient.

But who is fooling who? If this came out on the GMAT, I'm pretty sure that I would be somewhere else in time and space...

to explain. I'm not sure if I'm correct, but t0 = 3, using the equation you get the sequence: t1 = 4, t2=6, t3=9, t4=12... etc. From this, you should be able to see that 1 is insufficient (because you'll end up with both odd and even integers). 2. is sufficient you'll only end up with even integers.

Akamaibrah; please tell me if i'm completely off base. thank you.

Hallo guys, the way i understood the question t1=t0+n, t0 is 3 , now from a) seems that n could be 5 ,since n+1 should be divisible by 3, so by sustitution seems that t1=8, t2=13,t3=18 so we have both even and odds , which makes a) insufficient. For B seems that n could be 9 since n-1 is divisible by 4 so i get t1=3+9=12, t2=12+9=21, t3=30 or i get again odd and even numbers which makes statement B also insufficient..can you please tell me where my mistake is .. cause i see that most of youn think that B is sufficient ..thanks

tn=tn-1+n

t0=3
t1= 4
t2=6
t3=9
t4=13
t5=18
t6=24
t7=31
t8=39
t9=48
t10=58
t11=69
t12=81
t13=94

1) n+1 divisible by 3 (pick t2,t5,t8,t11 and you get both even and odd integers).

2) n-1 is divisible by 4 (pick t5,t9,t13 and you get only even numbers)
sufficient alone!

this was the quickest way for me to sort of guess what the right answer may be; although i may be completely wrong. [/b]

thanks titlelist , I thought that this n+1 and n-1 is valid for the n that participate in the sequence , and not for the numerator of t, so now it is clearer ..thanks

My pleasure! I hope I'm right.

Mr PHD:

While you are entitiled to your opinion, I think that your evaluation of the problem does not service to our readers and encourages them to "throw their hands up" when faced with a difficult problem. In reality, a well-designed problem seems difficult on the surface, but is readily solvable with the right approach and insight.

Granted, sometimes people post questions on this forum that are out-of=bounds for the GMAT. ETS and MGMAT, however, are very careful to only post questions that are solvable using basic mathematics -- once the student can identify the right approach(es). If you think that you need "advanced numerical theory" and "discrete analysis" and "inductive proofs", then you may solve the problem in an interesting manner, but you will have missed the point.

I was able to solve this problem in one minute using simple logic, properties of odd and even numbers, and understanding of what a comprises a set of numbers with a common divisor and remainder.

This is a fair problem. By approaching it as such, you will open your mind up to entertaining various approaches rather than being tempted to cry "foul" every time a challenging problem is in front of you. If you "pooh=pooh" it by assuming that you will never get such a problem on the GMAT, you will be unpleasantly surprised and unprepared.

BTW, some of you have approached this problem in a nice way, though the analysis could be a little more efficient. (Also, t2 = 6 not 4). Good job.

Yes! T2=6 - my bad - transcribing error!

If you look at the "n" values which result in even Tn values, they are:
1,2,5,6,9,10,13,14,17,18,20,21

which is the pattern 4n+1, 4n+2, 4n+1, 4n+2 repeated, where n >= 0.

So the choices are: I)either n-1 should be divisible by 4, or II) n-2 should be divisible by 4.

I) is the choice (2) in the question, so (B) is the answer.


Let me know if you think this is correct. It took me 4 mins to figure this out, and I doubt whether I will be able to do this on the GMAT under pressure ...

Akamai Brah,

I'll wait for the MGMAT solution before responding fully to your post, since you have talked about what you did, but not how and why you did it.

Cheers.

Nice solution. But I guess you can knock out the rest of the problems in 50 minutes, and have 25 minutes left over for the 750+ question .

I don't know what the "official" solution is but this is how i solved it.

OK, we are given a well-defined sequence, and we want to know whether the nth term is even (or odd -- doesn't matter). Since we only care about oddness or evenness (lets call this "parity" for lack of a better name), to save time we don't really have to add up actual numbers.

Make a table and let O = ODD and E = EVEN:

n=0, t0 = O
n=1, t1 = t0 + n = O + O = E
n=2, t2 = t1 + n = E + E = E
n=3, t3 = t2 + n = E + O = O
n=4, t4 = t3 + n = O + E = O
n=5, t5 = t4 + n = O + O = E
n=6, t6 = t5 + n = E + E = E
etc.

somewhere between the 5th and 9th term, one should be able to recognize that the pattern must repeat itself every 4 terms. Hence, from any initial term, every 4th term will have the same "parity".

1) this simply means that all n+1's are multiples of 3. Hence the set of all n's is simply each multiple of 3 minus 1, or every 3rd number starting with 2. Since every 3rd term is "out of sync" with the above pattern (which has period 4), one can quickly deduce that for any given possible n, we cannot tell whether tn is odd or even. NOT SUFF

2) all n-1's are multiples of 4. Hence all n's = every 4th number starting with 5. Since t5 is EVEN and every 4th number has same "parity", this is all we need to know. SUFF