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Is the positive integer x divisible by each integer from 2 through 6 ?
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14 May 2016, 03:47
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Is the positive integer x divisible by each integer from 2 through 6 ?
The question asks whether x is divisible by the LCM of 2, 3, 4, 5, and 6, which is 60.
(1) x = 10m, where m is a positive integer divisible by each integer from 2 through 5.
m is divisible by the LCM of of 2, 3, 4, and 5, which is 60. Thus x = 10*(multiple of 60). Sufficient.
(2) 10x = n, where n is a positive integer divisible by each integer from 2 through 9.
n is divisible by the LCM of of 2, 3, 4, 5, 6, 7, 8, and 9, which is 2520 (9*7*8*5). Thus 10x = (multiple of 2520) --> x = (multiple of 252). If x = 252, then the answer is NO but if x is say, 252*60, then the answer is YES. Not sufficient.
Re: Is the positive integer x divisible by each integer from 2 through 6 ?
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14 May 2016, 03:36
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pego2008 wrote:
Is the positive integer x divisible by each integer from 2 through 6 ?
1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5
2- 10x = n, where n is a positive integer divisible by each integer from 2 through 9
Thanks in advance
clearly A is suff.... 1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5 x=10m.. and m is a multiple of 2 to 5................ since m is a multiple of 2 and 3 both, it is also div by 6.. hence suff
2-10x = n, where n is a positive integer divisible by each integer from 2 through 9 10x = n...
if n = 2*3*4*5*6*7*8*9... 10x = 2*3*4*5*6*7*8*9... so x = 3*4*6*7*8*9... so x need not be div by 5...NO
Is the positive integer x divisible by each integer from 2 through 6 ?
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07 Jul 2016, 12:16
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considering the numbers one at a time in order to figure out what x will need, at a minimum, in it's prime factorization:
2: in order for x to be divisible by 2, x needs a 2 in its PF 3: in order for x to be divisible by 3, x needs a 3 in its PF (so now we have 2*3) 4: in order for x to be divisible by 4, x needs another 2 in its PF (so now we have 2*2*3) 5: in order for x to be divisible by 5, x needs a 5 in its PF (so now we have 2*2*3*5) 6: in order for x to be divisible by 6, x needs a 2 and a 3 in its PF (we already have that so our min doesn't change; x still needs, at min, 2*2*3*5 in its PF)
So we need enough info to prove that x has at least 2*2*3*5 in its PF
Statement 1.
x = (10)(m) ---> rewriting m in terms of what we know about its PF --> x=(2*5)(2*3*2*5*whatever else might be in m)
We can see right away that x has at least 2*2*3*5 in its PF. Sufficient.
Statement 2
10x = n ---> x = n/10 --> rewriting m in terms of what we know about its PF, and factoring 10 --> x = (2*2*2*3*3*5*7*whatever else might be in n)/2*5 --> the two and 5 cancel to get ---> x = (2*2*3*3*7*whatever else might be in n)
We don't know if x has 2*2*3*5 in its PF because we don't know for sure that there is a 5; there may or may not be a 5 in the 'whatever else might be in n' part. Insufficient.
Re: Is the positive integer x divisible by each integer from 2 through 6 ?
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08 Jul 2016, 22:04
If we modify the original condition and the question, since x has to be divisible by 2, 3, 4, 5 and 6, the question becomes “is it x a multiple of 60?” In case of con 1), from x=10m, m is divisible by 2, 3, 4 and 5. Hence, x is a multiple of 60. The answer is yes and the condition is sufficient. Thus, the correct answer is A.
- Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.
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Is the positive integer x divisible by each integer from 2 through 6 ?
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23 Jul 2016, 13:19
chetan2u wrote:
pego2008 wrote:
Is the positive integer x divisible by each integer from 2 through 6 ?
1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5
2- 10x = n, where n is a positive integer divisible by each integer from 2 through 9
Thanks in advance
clearly A is suff.... 1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5 x=10m.. and m is a multiple of 2 to 5................ since m is a multiple of 2 and 3 both, it is also div by 6.. hence suff
2-10x = n, where n is a positive integer divisible by each integer from 2 through 9 10x = n...
if n = 2*3*4*5*6*7*8*9... 10x = 2*3*4*5*6*7*8*9... so x = 3*4*6*7*8*9... so x need not be div by 5...NO
if n = 2*3*4*5*6*7*8*9*5... ans is YES Insuff
A
Hello, can someone please explain to me why can we add another five in the second statement? It's kind of driving me nuts!
Re: Is the positive integer x divisible by each integer from 2 through 6 ?
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23 Jul 2016, 20:17
1
Avigano wrote:
chetan2u wrote:
pego2008 wrote:
Is the positive integer x divisible by each integer from 2 through 6 ?
1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5
2- 10x = n, where n is a positive integer divisible by each integer from 2 through 9
Thanks in advance
clearly A is suff.... 1- x = 10m, where m is a positive integer divisible by each integer from 2 through 5 x=10m.. and m is a multiple of 2 to 5................ since m is a multiple of 2 and 3 both, it is also div by 6.. hence suff
2-10x = n, where n is a positive integer divisible by each integer from 2 through 9 10x = n...
if n = 2*3*4*5*6*7*8*9... 10x = 2*3*4*5*6*7*8*9... so x = 3*4*6*7*8*9... so x need not be div by 5...NO
if n = 2*3*4*5*6*7*8*9*5... ans is YES Insuff
A
Hello, can someone please explain to me why can we add another five in the second statement? It's kind of driving me nuts!
Hi,
St2: 10x = n --> x = n/10 n can be any integer. If n = 2*3*4*5*6*7*8*9*5, then x is divisible by each integer from 2 through 6 --> Another 5 is added here to show sufficiency. If n = 2*3*4*5*6*7*8*9, then x is not divisible by 5 because n/10 consumes a 2 and 5 --> In this case St2 is not sufficient We can conclude that St2 is not sufficient at all times.
Re: Is the positive integer x divisible by each integer from 2 through 6 ?
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23 Dec 2017, 01:26
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