C is again eliminated since we cannot get a valid value of m.Night reader wrote:Is the sum of all factors of a positive integer m less than 13131313450?
(1) m < 13131313450
(2) m > 13131313450
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.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.
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.anshumishra wrote: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.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.
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.
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.Night reader wrote: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.anshumishra wrote: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.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.
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.
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.Night reader wrote: The sum of factors for 67 is equal to 67 not 64.
yes it's 68, and the concept demanded followsanshumishra wrote: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.Night reader wrote: The sum of factors for 67 is equal to 67 not 64.
Cool. I used the specialized case of the formula you mentioned (by restricting my test case to use only 2).Night reader wrote:yes it's 68, and the concept demanded followsanshumishra wrote: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.Night reader wrote: The sum of factors for 67 is equal to 67 not 64.
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
the original question was posted at GMAT club web-site; I have changed the problem's content.tomada wrote:I'm curious, what is the source of this question?
it's easy to get a "No" answer -- just take any small number, for which the sum of factors will also be small.Night reader wrote:Is the sum of all factors of a positive integer m less than 13,131,313,450?
(1) 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).(2) m > 13131313450
