If n = t^3 for some positive integer t and if 8, 9 and 10 are each factors of n, which of the following must be a factor of n?
A. 16
B. 81
C. 175
D. 225
E. 275
A. 16
B. 81
C. 175
D. 225
E. 275
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If n = t^3 for some positive integer
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A lot of integer property questions can be solved using prime factorization.ritumaheshwari02 wrote:If n = t^3 for some positive integer t and if 8, 9 and 10 are each factors of n, which of the following must be a factor of n?
A. 16
B. 81
C. 175
D. 225
E. 275
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is divisible by k (i.e., k is a factor of N), then k is "hiding" within the prime factorization of N
Examples:
24 is divisible by 3 <--> 24 = 2x2x2x3
70 is divisible by 5 <--> 70 = 2x5x7
330 is divisible by 6 <--> 330 = 2x3x5x11
56 is divisible by 8 <--> 56 = 2x2x2x7
Okay, onto the question.
We're told that 8 is a factor of n, which means 8 must be hiding in the prime factorization of n.
In other words, we know that n = 2x2x2x?x?x?... (the ?'s represent other values that could also be in the prime factorization of n)
IMPORTANT: Since n = (t)(t)(t), we can conclude that 2 must be a factor of t.
Similarly, told that 9 is a factor of n, which means 9 must be hiding in the prime factorization of n.
In other words, we know that n = 3x3x?x?x?...
IMPORTANT: Since n = (t)(t)(t), we can conclude that 3 must be a factor of t.
Using similar logic, we can conclude that 5 must be a factor of t.
So, if 2, 3 and 5 are all factors of t, we know that t = 2x3x5x?x?x?...
Since n = t^3, we know that n = (2x3x5x?x?x?)(2x3x5x?x?x?)(2x3x5x?x?x?)
In other words, n = 2x2x2x3x3x3x5x5x5x?x?x?x?x?x?...
From here, it's easy to see what must be a factor of n.
A) Since we can only be certain that n has three 2's hiding in its prime factorization, we cannot conclude that 16 is a factor of n
B) Since we can only be certain that n has three 3's hiding in its prime factorization, we cannot conclude that 81 is a factor of n
C) 175 = 5x5x7. Since we 7 may or may not be hiding in the prime factorization of n, we cannot conclude that 175 is a factor of n
D) 225 = 3x3x5x5. Since there are two 3's and two 5's hiding in the prime factorization of n, we can conclude that 225 is a factor of n
E) 275 = 5x5x11. Since we 11 may or may not be hiding in the prime factorization of n, we cannot conclude that 275 is a factor of n
The correct answer is D
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Brent
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Brent@GMATPrepNow
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Simple Way :
N has to be LCM of 8,9 and 10
LCM of 8,9 and 10 - 360
Factors of 360 is 2^3 * 3^2 * 5^1
We know that n is a cube of a number and 360 is not a cube
So we multiply 360 * 3^1 * 5^2 = 27000 (to make it a perfect cube)
Divide 27000 by the options, the one that completely divides it is a factor
N has to be LCM of 8,9 and 10
LCM of 8,9 and 10 - 360
Factors of 360 is 2^3 * 3^2 * 5^1
We know that n is a cube of a number and 360 is not a cube
So we multiply 360 * 3^1 * 5^2 = 27000 (to make it a perfect cube)
Divide 27000 by the options, the one that completely divides it is a factor
Nice shortcut 
ritind wrote:Simple Way :
N has to be LCM of 8,9 and 10
LCM of 8,9 and 10 - 360
Factors of 360 is 2^3 * 3^2 * 5^1
We know that n is a cube of a number and 360 is not a cube
So we multiply 360 * 3^1 * 5^2 = 27000 (to make it a perfect cube)
Divide 27000 by the options, the one that completely divides it is a factor
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The Iceman
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n=k*LCM(8,9,10)=360 k = (2^3)*(3^2)*5*k and k = 3*(5^2)*z= 75*zritumaheshwari02 wrote:If n = t^3 for some positive integer t and if 8, 9 and 10 are each factors of n, which of the following must be a factor of n?
A. 16
B. 81
C. 175
D. 225
E. 275
Only option D is divisible by 75, and hence the answer
Hii,
i didnt understand y 225 is a factor.. n has 2,3 and 5 as basic factors but 225 doesnt have 2 as its factor..please explain
i didnt understand y 225 is a factor.. n has 2,3 and 5 as basic factors but 225 doesnt have 2 as its factor..please explain
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GMATGuruNY
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Since the correct answer choice MUST be a factor of n, the correct answer choice must be able to divide evenly into the LEAST POSSIBLE VALUE of n.ritumaheshwari02 wrote:If n = t^3 for some positive integer t and if 8, 9 and 10 are each factors of n, which of the following must be a factor of n?
A. 16
B. 81
C. 175
D. 225
E. 275
n = t³, where t is a positive integer.
Implication:
If prime number p is a factor of n, then p³ is a factor of n.
Since 8, 9 and 10 are all factors of n, prime numbers 2, 3, and 5 are all factors of n, implying that 2³, 3³, and 5³ must all be factors of n.
Thus, the LEAST POSSIBLE VALUE of n = 2³3³5³.
Of the 5 answer choices, only 225 divides evenly into the least possible value of n:
2³3³5³/225 = 2³3³5³/3²5² = 2³*3*5.
The correct answer is D.
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Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
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An easy way of thinking about this: any cube has three equal roots. Any prime factor of the cube must be part of EACH ROOT, so a cube must have THREE of each of its prime factors.Vardhamanl wrote:Hii,
i didnt understand y 225 is a factor.. n has 2,3 and 5 as basic factors but 225 doesnt have 2 as its factor..please explain
For instance, 15³ = 3³ * 5³; 49³ = 7³ * 7³; 100³ = 2³ * 5³, etc.
Since 5 is a factor of n, 5³ is also a factor of n. Similarly, since 3 is a factor of n, 3³ is a factor of n. 225 is a factor of 5³ * 3³, so 225 must be a factor of n.
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Jim@StratusPrep
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We know that 2³3³5³ is the smallest possible value of n. The cube of an integer will have a prime factorization that contains exponents with multiples of 3 and since none of the factors of n have more than 3 of any individual prime factor, we know 3 is the smallest any of the exponents can be.
From there, just look for a number that is divisible into 2³3³5³
A and B have more 2s and 3s respectively
C and E have a 7 and 11 respectively
D
From there, just look for a number that is divisible into 2³3³5³
A and B have more 2s and 3s respectively
C and E have a 7 and 11 respectively
D
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