DS number theory? (2)

Expert replies

Previous Topic Next Topic

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Thu Dec 30, 2010 1:52 pm

Is the sum of all factors of a positive integer m less than 13131313450?
(1) m < 13131313450
(2) m > 13131313450

Like this post

Source: Beat The GMAT — Data Sufficiency | Thu Dec 30, 2010 1:52 pm

ragz: Junior | Next Rank: 30 Posts; Posts: 10; Joined: Wed Dec 29, 2010 10:24 pm; Thanked: 3 times

by ragz » Thu Dec 30, 2010 1:59 pm

Night reader wrote:Is the sum of all factors of a positive integer m less than 13131313450?
(1) m < 13131313450
(2) m > 13131313450

C is again eliminated since we cannot get a valid value of m.

For sentence 1, let's always try taking the biggest value less than the number mentioned in the inequation which 13131313450. At a minimum, any number will have 1 and the number itself as its factors. Hence, if you sum up the factors, it will always be greater than the number itself. In this case, let's take m=13131313449, we have atleast 1 and 13131313449 apart from other factors. So, the sum will be 13131313450 + sum of other factors which is greater than or equal to 13131313450. If we take a smaller number, like 20, then the sum of the factors will be definitely less than 13131313450. So, sentence 1 alone is not sufficient.

Hence, A and D can be eliminated.

For Sentence 2, based on the explanation in the earlier post, for any value of m, the sum of factors will be greater than 13131313450.

Hence, B is the answer.

Hope this helps! I would appreciate if someone else can validate my approach.

--
Ravi.

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Thu Dec 30, 2010 2:05 pm

that's a way to solve DS
to add we are missing the concept of finding the sum of all factors for a number here; your solution is sound, as it's based on elimination of impossible choices out of given. Yet we are left without application of sum of all factors for the number here. Never mind, your answer was correct.

Like this post

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Thu Dec 30, 2010 2:37 pm

Night reader wrote:that's a way to solve DS
to add we are missing the concept of finding the sum of all factors for a number here; your solution is sound, as it's based on elimination of impossible choices out of given. Yet we are left without application of sum of all factors for the number here. Never mind, your answer was correct.

Not sure whether you are looking for the exact same thing. Let me add one more concept which came to my mind when I saw the problem.

Suppose we have any number (say 67), Find the number which is less than that number and a power of 2, here it is
2^6 = 64

Sum of factor = (2^7-1)/2-1 = 2^7 - 1 = 127 (One less than the next power of 2)
So, we are always going to have at least one number which has the sum of prime factors just one less than the next power of 2 (here 7).

Does that makes sense ? Not sure if this was useful, however thought to share.

Thanks
Anshu

(Every mistake is a lesson learned )

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Thu Dec 30, 2010 3:48 pm

anshumishra wrote:
Night reader wrote:that's a way to solve DS
to add we are missing the concept of finding the sum of all factors for a number here; your solution is sound, as it's based on elimination of impossible choices out of given. Yet we are left without application of sum of all factors for the number here. Never mind, your answer was correct.
Not sure whether you are looking for the exact same thing. Let me add one more concept which came to my mind when I saw the problem.

Suppose we have any number (say 67), Find the number which is less than that number and a power of 2, here it is
2^6 = 64

Sum of factor = (2^7-1)/2-1 = 2^7 - 1 = 127 (One less than the next power of 2)
So, we are always going to have at least one number which has the sum of prime factors just one less than the next power of 2 (here 7).

Does that makes sense ? Not sure if this was useful, however thought to share.

this makes sense when prime factorization of the number is equal to 2^n; when the primes other than 2 are involved the difference varies. The sum of factors for 67 is equal to 67 not 64.

Like this post

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Thu Dec 30, 2010 3:52 pm

Night reader wrote:
anshumishra wrote:
Night reader wrote:that's a way to solve DS
to add we are missing the concept of finding the sum of all factors for a number here; your solution is sound, as it's based on elimination of impossible choices out of given. Yet we are left without application of sum of all factors for the number here. Never mind, your answer was correct.
Not sure whether you are looking for the exact same thing. Let me add one more concept which came to my mind when I saw the problem.

Suppose we have any number (say 67), Find the number which is less than that number and a power of 2, here it is
2^6 = 64

Sum of factor = (2^7-1)/2-1 = 2^7 - 1 = 127 (One less than the next power of 2)
So, we are always going to have at least one number which has the sum of prime factors just one less than the next power of 2 (here 7).

Does that makes sense ? Not sure if this was useful, however thought to share.
this makes sense when prime factorization of the number is equal to 2^n; when the primes other than 2 are involved the difference varies.

Agree. This was just an alternate proof (based on the sum of factorization technique ) that there is always some number less than any number "n" which has the sum of its factors greater than n.

Thanks
Anshu

(Every mistake is a lesson learned )

Like this post

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Thu Dec 30, 2010 3:57 pm

Night reader wrote: The sum of factors for 67 is equal to 67 not 64.

Yes, it is actually 67+1 = 68. The question was on the line of for all <=67, if the sum of factors is less than 67? So, I replied accordingly.

Thanks
Anshu

(Every mistake is a lesson learned )

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Thu Dec 30, 2010 4:10 pm

anshumishra wrote:
Night reader wrote: The sum of factors for 67 is equal to 67 not 64.
Yes, it is actually 67+1 = 68. The question was on the line of for all <=67, if the sum of factors is less than 67. So, I replied accordingly.

yes it's 68, and the concept demanded follows

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The sum of factors of n will be expressed by the formula: {(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)} fract/ {(a-1)*(b-1)*(c-1)}

Example: Finding the sum of all factors of 450: 450=2^1*3^2*5^2

The sum of all factors of 450 is {(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)} fract/ {(2-1)*(3-1)*(5-1)}={3*26*124} fract/ {1*2*4}=1209

Like this post

anshumishra: Legendary Member; Posts: 543; Joined: Tue Jun 15, 2010 7:01 pmFollowed by: 3 members; Thanked: 147 times

by anshumishra » Thu Dec 30, 2010 4:16 pm

Night reader wrote:
anshumishra wrote:
Night reader wrote: The sum of factors for 67 is equal to 67 not 64.
Yes, it is actually 67+1 = 68. The question was on the line of for all <=67, if the sum of factors is less than 67. So, I replied accordingly.
yes it's 68, and the concept demanded follows

First make prime factorization of an integer n=a^p*b^q*c^r, where a, b, and c are prime factors of n and p, q, and r are their powers.

The sum of factors of n will be expressed by the formula: {(a^{p+1}-1)*(b^{q+1}-1)*(c^{r+1}-1)} fract/ {(a-1)*(b-1)*(c-1)}

Example: Finding the sum of all factors of 450: 450=2^1*3^2*5^2

The sum of all factors of 450 is {(2^{1+1}-1)*(3^{2+1}-1)*(5^{2+1}-1)} fract/ {(2-1)*(3-1)*(5-1)}={3*26*124} fract/ {1*2*4}=1209

Cool. I used the specialized case of the formula you mentioned (by restricting my test case to use only 2).
Good that you put it here; this could be useful for people here. Thanks Night reader !

Thanks
Anshu

(Every mistake is a lesson learned )

Like this post

tomada: Master | Next Rank: 500 Posts; Posts: 406; Joined: Mon Jan 25, 2010 11:36 amFollowed by: 4 members; Location: Syracuse, NY; Thanked: 23 times; GMAT Score:740

by tomada » Fri Dec 31, 2010 11:52 am

I'm curious, what is the source of this question?

I'm really old, but I'll never be too old to become more educated.

Like this post

Night reader: Legendary Member; Posts: 1337; Joined: Sat Dec 27, 2008 6:29 pmFollowed by: 10 members; Thanked: 127 times

by Night reader » Fri Dec 31, 2010 5:10 pm

tomada wrote:I'm curious, what is the source of this question?

the original question was posted at GMAT club web-site; I have changed the problem's content.

Like this post

GMAT/MBA Expert

lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Sun Jan 02, 2011 12:37 am

some of the solutions given here are a bit on the over-complicated side; note that you don't have to find the sum of the factors -- you just have to determine whether it's less than this weird value (13,131,313,450).

Night reader wrote:Is the sum of all factors of a positive integer m less than 13,131,313,450?
(1) m < 13131313450

it's easy to get a "No" answer -- just take any small number, for which the sum of factors will also be small.
to get a "yes" answer, just pick any even number that's decently close to 13,131,313,450. then, as factors, you'll have (among many, many others) that number itself, and also half that number; if your original number is close enough to 13,131,313,450, those two values alone will sum to more than that number.
for instance, if you take 10,000,000,000, then you have the two factors 10,000,000,000 and 5,000,000,000, whose sum is already more than enough.
insufficient.

(2) m > 13131313450

m is always a factor of itself, so, in these cases, that value alone is enough to figure out that the sum of factors is more than the reference number (since m itself is more than the reference number).

Ron has been teaching various standardized tests for 20 years.

--

Pueden hacerle preguntas a Ron en castellano
Potete chiedere domande a Ron in italiano
On peut poser des questions Ã Ron en franÃ§ais
Voit esittÃ¤Ã¤ kysymyksiÃ¤ Ron:lle myÃ¶s suomeksi

--

Quand on se sent bien dans un vÃªtement, tout peut arriver. Un bon vÃªtement, c'est un passeport pour le bonheur.

Yves Saint-Laurent

--

Learn more about ron

Like this post

GMAT/MBA Expert

lunarpower: GMAT Instructor; Posts: 3380; Joined: Mon Mar 03, 2008 1:20 amFollowed by: 1535 members; Thanked: 2256 times; GMAT Score:800

by lunarpower » Sun Jan 02, 2011 12:49 am

oh, and, also, note that this couldn't be an official problem -- the 2 statements are obviously contradictory.

the test NEVER has two contradictory statements as (1) and (2) -- there is always a non-empty overlap between the two statements (i.e., there is at least one number that satisfies both of them). note that this is still the case even when the answer to the problem is (a), (b), or (d), the cases when it's never actually necessary to combine the statements.

(it's unlikely that knowing this convention will actually help you solve any problems, but at least it can help you spot problems from sketchy/unreliable sources.)

Like this post

Post new topic Post Reply

• Page 1 of 1

GMAT Course Reviews

560 reviews

1168 reviews

431 reviews

359 reviews

147 reviews

5/5

5 Star (560 Reviews)

"Target Test Prep is the closest to the official version of the GMAT exam, about 99% accuracy in terms of the quality and quantity of information. The course has excellently created singular sets of focused lessons and tests for every possible topic that one could come across in the official GMAT exam."

"The TTP course maximizes the efficiency of the time you spend studying. It will take time and effort but I could almost guarantee that if you complete the course exactly as it is laid out you will get an amazing score. They also have a very responsive team willing to help with any questions you might have."

"TTP has two things that I think no other test prep company offers: A teaching approach that reinforces understanding and an attitude that will give you the mental preparedness needed to succeed on the test. TTP gives you a deep understanding of the concept you need to know while teaching you how to think."

Write a Review