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For any integer k > 1, the term “length of an integer” [#permalink]

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22 Jan 2011, 12:31

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For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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22 Jan 2011, 12:49

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ajit257 wrote:

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5 b. 6 c. 15 d. 16 e. 18

Can some explain an elegant way of doing such a problem which would take less time.

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

A 5 B 6 C 15 D 16 E 18

Merging similar topics. Ask if anything remains unclear.

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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15 Mar 2012, 06:15

Bunuel wrote:

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel, I tried solving this question. However, I thought x and y to be different. That's why I put x= 2 and y = 3, in order to minimize the prime bases and thus maximize the powers of the primes. Isnt the question implying x and y to be different. Otherwise the given equation x+3y<1000 is as good as x+3x<1000

..getting what I am trying to put across? _________________

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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15 Mar 2012, 07:33

Expert's post

imhimanshu wrote:

Bunuel wrote:

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel, I tried solving this question. However, I thought x and y to be different. That's why I put x= 2 and y = 3, in order to minimize the prime bases and thus maximize the powers of the primes. Isnt the question implying x and y to be different. Otherwise the given equation x+3y<1000 is as good as x+3x<1000

..getting what I am trying to put across?

We are not told that x and y are distinct, so we cannot assume this. _________________

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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15 May 2012, 10:27

Bunuel wrote:

imhimanshu wrote:

Bunuel wrote:

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel, I tried solving this question. However, I thought x and y to be different. That's why I put x= 2 and y = 3, in order to minimize the prime bases and thus maximize the powers of the primes. Isnt the question implying x and y to be different. Otherwise the given equation x+3y<1000 is as good as x+3x<1000

..getting what I am trying to put across?

We are not told that x and y are distinct, so we can not assume this. Next, even if we were told that they are distinct the answer still would be D: 2^8*3=768<100 also has the length of 8+1=9.

Hi Bunuel

x+3y<1000 and if x and y are distinct then shouldn't this be 2^9+3^5<1000 for a length of 14 which is not among the options, so I guess the question didn't mean that they are distinct. Please can you explain once more if they are distinct how can the answer still be 16 (D)

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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16 May 2012, 02:12

Expert's post

Joy111 wrote:

Hi Bunuel

x+3y<1000 and if x and y are distinct then shouldn't this be 2^9+3^5<1000 for a length of 14 which is not among the options, so I guess the question didn't mean that they are distinct. Please can you explain once more if they are distinct how can the answer still be 16 (D)

Thanks

We are not told that x and y are distinct. But if we were told that, the answer would be 13 not 14: x+3y=2^9+3*3^4=755<1,000 --> 9+4=13. _________________

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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16 May 2012, 02:19

Bunuel wrote:

Joy111 wrote:

Hi Bunuel

x+3y<1000 and if x and y are distinct then shouldn't this be 2^9+3^5<1000 for a length of 14 which is not among the options, so I guess the question didn't mean that they are distinct. Please can you explain once more if they are distinct how can the answer still be 16 (D)

Thanks

We are not told that x and y are distinct. But if we were told that, the answer would be 13 not 14: x+3y=2^9+3*3^4=755<1,000 --> 9+4=13.

Hi Bunuel

shouldn't we take 3*3^4= 3^5 ( adding the exponents with same base) and hence 9+5= 14 ?

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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16 May 2012, 02:22

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Expert's post

Joy111 wrote:

Bunuel wrote:

Joy111 wrote:

Hi Bunuel

x+3y<1000 and if x and y are distinct then shouldn't this be 2^9+3^5<1000 for a length of 14 which is not among the options, so I guess the question didn't mean that they are distinct. Please can you explain once more if they are distinct how can the answer still be 16 (D)

Thanks

We are not told that x and y are distinct. But if we were told that, the answer would be 13 not 14: x+3y=2^9+3*3^4=755<1,000 --> 9+4=13.

Hi Bunuel

shouldn't we take 3*3^4= 3^5 ( adding the exponents with same base) and hence 9+5= 14 ?

No. Please read the question carefully "what is the maximum possible sum of the length of x and the length of y". The length of x is 9 and the length of y is 4, so the sum is 9+4=13. _________________

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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16 May 2012, 04:18

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We are not told that x and y are distinct. But if we were told that, the answer would be 13 not 14: x+3y=2^9+3*3^4=755<1,000 --> 9+4=13.[/quote]

Hi Bunuel

shouldn't we take 3*3^4= 3^5 ( adding the exponents with same base) and hence 9+5= 14 ?[/quote]

No. Please read the question carefully "what is the maximum possible sum of the length of x and the length of y". The length of x is 9 and the length of y is 4, so the sum is 9+4=13.[/quote]

Awesome , thanks ,really fell in the trap for that one !! + 1

here is my approach : we know that : x > 1, y > 1, and x + 3y < 1000, and it is given that length means no of factors. for any value of x and y, the max no of factors can be obtained only if factor is smallest no & all factors are equal. hence, lets start with smallest no 2. 2^1 =2 2^2 =4 2^3=8 2^4=16 2^5=32 2^6=64 2^7=128 2^8=256 2^9=512 2^10 =1024 (opps//it exceeds 1000, so, x can't be 2^10) so, max value that X can take is 2^9 , for which has "length of integer" is 9. now, since x =512 , & x+3y<1000 so, 3y<488 ==> y<162 so, y can take any value which is less than 162. and to get the maximum no of factors of smallest integer, we can say y=2^7 for 2^7 has length of integer is 7.

SO, combined together : 9+7 = 16. And is D.

Hope it will help. _________________

kudos me if you like my post.

Attitude determine everything. all the best and God bless you.

What's the source of this problem? Basically, you know that in order to get the longest "length", we'll want all 2's in the prime factorization. So how can we satisfy x + 3y < 1000 where x and y are both 2 to the nth power.

TOO HIGH, but that's interesting. It definitely looks like we can play around with this somehow to reach it. We know that it won't make sense to increase to 2^9 for y's value because 2^9 is 512 and 512*3 > 1000. How about the other way around?

2^9 + 3(2^7) = ? 512 + 3(128) = 896

9 + 7 = 16.

We know we won't be able to get much higher than that because 2^10 as x is the only other move we could try and that is > 1000 by itself. (It's 1024 which you should have memorized!) So answer is D = 16.

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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01 Aug 2013, 07:23

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Bunuel wrote:

ajit257 wrote:

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5 b. 6 c. 15 d. 16 e. 18

Can some explain an elegant way of doing such a problem which would take less time.

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel, Don't we need to check for the other case. i.e when try to maximise the length of Y rather than that of X??

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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29 Aug 2013, 20:37

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12bhang wrote:

Bunuel wrote:

ajit257 wrote:

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5 b. 6 c. 15 d. 16 e. 18

Can some explain an elegant way of doing such a problem which would take less time.

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel, Don't we need to check for the other case. i.e when try to maximise the length of Y rather than that of X??

I think we can try both cases to see which gives you the maximum Length.

1) maximize Y

x + 3 y < 1000 2^9 = 512 2 ^8 = 256 so y = 2^ 8 and now x is 999 - 3(256) = 231 x can be 2^7 = 128 So total length = 8 + 7 = 15

2) maximize x

x + 3 y < 1000 2^9 = 512 2 ^ 10 = 1024 length of x = 9 now, y = (999 - 512 ) / 3 = 162.x or y is 2^7

Total length is 9+7 = 16.

Now, you might want to NOT do the maximization of y because you know that most 2s will be in x and NOT y. For example, we have to first realize that we want the number with the MOST 2s in it. x can be that number as illustrated by this example: If y = 512 -> 512 * 3 > 1000 and if x = 512 , well x can be 512. So, we would get a length of 9 out of it.

If we were doing, x + y < 1000 then the lengths can be inter-changed; however, because of the 3 next to y, we know that the length of y will have to be LESS than the length of x.

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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11 Nov 2013, 10:17

Bunuel wrote:

ajit257 wrote:

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5 b. 6 c. 15 d. 16 e. 18

Can some explain an elegant way of doing such a problem which would take less time.

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Hi Bunuel,

How did you know to start with "x" and not "y"? What is the logic? When I start with "y" I get a different result....

Re: For any integer k > 1, the term “length of an integer” [#permalink]

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20 Nov 2014, 02:03

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Re: For any integer k > 1, the term “length of an integer” [#permalink]

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04 Sep 2015, 07:35

Bunuel wrote:

ajit257 wrote:

For any integer k > 1, the term “length of an integer” refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 × 2 × 2 × 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y < 1000, what is the maximum possible sum of the length of x and the length of y?

a. 5 b. 6 c. 15 d. 16 e. 18

Can some explain an elegant way of doing such a problem which would take less time.

Basically the length of an integer is the sum of the powers of its prime factors. For example the length of 24 is 4 because 24=2^3*3^1 --> 3+1=4.

Given: x+3y<1,000. Now, to maximize the length of x or y (to maximize the sum of the powers of their primes) we should minimize their prime bases. Minimum prime base is 2: so if x=2^9=512 then its length is 9 --> 512+3y<1,000 --> y<162.7 --> maximum length of y can be 7 as 2^7=128 --> 9+7=16.

Answer: D.

Bunuel, i must say, u rock man!!!

gmatclubot

Re: For any integer k > 1, the term “length of an integer”
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04 Sep 2015, 07:35

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