a,b and c are positive integers such that abc= 450 and

I will go like this

450 = ax thus a = 450/x = intiger x has 3*3*2 possibilities

least is 1 and max is 450 and thus A max is 450 AND MINIMUM IS 1

A BELONGS TO { 1, 450, 3,9,5,25,2,6,,10,15,18,90,50,75,225,30,150,60}

,
same goes for b,c

but

a+4<b+2<c

IF A = 1 "LEAST POSSIBLE VALUE ", B-2 HAS TO BE > 1 THUS B COULD BE AT LEAST FIVE AND THE LEAST POSSIBLE VALUE OF C IN THIS CASE IS 6

A+B+C = 1+5+6 = 12

IF C IS MAX = 450 THUS B COULD BE MAX = 225 AND THUS ONE COULD BE MAX = 150


i DONT KNOW WHERE I AM GOING WRONG

kEVIN HELP ....PLZ

You aren't going wrong- the question was badly written- it has been corrected- good work yeez!

I'm getting a max range of 68 which is not in the answer choice, don't know where I'm wrong though

The minimum value for a+b+c that I'm coming up with is 28 for a as 3, b as 10 and c as 15.

The maximum value is 96 for a as 1, b as 5 and c as 90.

What is wrong with this approach?

Your approach is fine. I've had a lot of trouble writing this question properly. I had better take a break!

Ah yes...thanks Yezz. Can you give the OA and explanation kevincan?

We are told that a,b,c are integers abc=450 and that a+2<b<c-2, In other words a<b<c and at least two integers are between a and b and at least integers are between b and c.

How can we maximize a+b+c? By having a,b, and c as disperse as possible, so that c is really big. 450=2*3^2*5^2, so if a=1, b=5, c=90 we get a+b+c=96.

How can we mimimize a+b+c? By having a,b and c as close together as possible!

Can a=5? b*c would then be 2*3^2*5 The lowest possible value of b (remember b>a+2) would be 9. However in that case c would be 10, which is not possible, as c>b+2.

Can a=3? b*c would then be 2*3*5^2 b could be 6 and c would then be 25 (a+b+c=34) or b could be 10 and c would then be 15 (a+b+c=28).

If a=2 , b*c=3^2*5^2, every possible (b,c) for a=2 would yield a value of a+b+c>28.

So, the mimimum and maximum values of a+b+c will be 28 and 96 respectively and so the range of T is 68. Sorry for the trouble I had in getting the question written properly, but I think it was worth the wait. I daresay this is one of my better questions. Do you agree?

It was a good question. I enjoyed solving this one just as I enjoy solving some of your other questions. Are you writing these yourself?

Yes. They're for a book I'm putting together, albeit slowly. Contributions are most welcome, as my creativity comes in spurts. I have 250 questions so far, and I´d like 500 to make a really good book. Do you think such a book would have a market?

Yes, I think so especially if explanations are clear within it, if concepts are well isolated and applicable to real GMAT questions and if questions are ordered in level of difficulties like inside the OG.

An hard part is how to market it Such new book needs to have enough of credibility to compete other big names After all, why not trough GMATClub members... It could be

For the book to sell, it would have to be relevant to a GMAT test taker. So if conceptually it covers most of the things that are tested on the GMAT then it would do well. For mass appeal it should probably have a graded level of difficulty, starting from easier problems and then moving on to tougher ones.

Just my two cents

Need not be a GMAT book at all! These are lovely questions, and there are lots of math freaks and students out there.

Heck, I'd buy one for my kid

Luv your problems. And let us know when the book is out.

Graded difficulty is a double-edged sword. I sometimes think that prior knowledge of the difficulty conditions how we think (especially for DS). Yet for others, it helps to use their time more wisely.