If n is a positive integer and n2 is divisible by 72, then the largest positive integer that must divide n is:
a)6
b)12
c)24
d)36
e)48
answer later
Divisibility Quant Review (#169)
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Picking Number Approach:DCS80 wrote:If n is a positive integer and n2 is divisible by 72, then the largest positive integer that must divide n is:
a)6
b)12
c)24
d)36
e)48
answer later
Least possible value of n² such that n² is divisible by 72 is 72*2 = 144
Hence, minimum possible value of n = 12.
Largest possible integer that divides n is 12.
Algebraic Approach:
n² is divisible by 72
Hence we can write n² as 72k, where k is an positive integer.
Now, n = √n² = √(72k) = √[(2)*(36)*k] = 6√(2k)
Now for n to be an integer, k must be an even multiple of a perfect square.
Hence, we can write k = 2m², where is a positive integer.
Now, n = 6√(2k) = 6√(2*2*m²) = 12m
Hence, largest possible integer that divides n is 12
The correct answer is B.
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Since n² is the square of an integer, n² is a perfect square.DCS80 wrote:If n is a positive integer and n2 is divisible by 72, then the largest positive integer that must divide n is:
a)6
b)12
c)24
d)36
e)48
Since n² is divisible by 72, n² must be a perfect square greater than 72, implying the following options:
n² = 9*9 = 81. Not a multiple of 72.
n² = 10*10 = 100. Not a multiple of 72.
n² = 11*11 = 121. Not a multiple of 72.
n² = 12*12 = 144, which is a multiple of 72.
Thus, n=12 is the SMALLEST POSSIBLE VALUE of n.
Of the five answer choices, the greatest option that MUST divide into n=12 is B.
The correct answer is B.
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another way
n*n/72 gives positive int
n*n/2x2x2x3x3
so the each n must have 2x2x3 in order to give int
hence 12 is the greatest multiple of n
n*n/72 gives positive int
n*n/2x2x2x3x3
so the each n must have 2x2x3 in order to give int
hence 12 is the greatest multiple of n
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We are given that n^2/72 = integer or (n^2)/(2^3)(3^2) = integer.DCS80 wrote:If n is a positive integer and n2 is divisible by 72, then the largest positive integer that must divide n is:
a)6
b)12
c)24
d)36
e)48
However, since n^2 is a perfect square, we need to make 72 or (2^3)(3^2) a perfect square. Since all perfect squares consist of unique primes, each raised to an even exponent, the smallest perfect square that divides into n^2 is (2^4)(3^2) = 144.
Since n^2/144 = integer, then n/12 = integer, and thus the largest positive integer that must divide n is 12.
Answer: B
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---------------ASIDE #1--------------------------------------DCS80 wrote:If n is a positive integer and n2 is divisible by 72, then the largest positive integer that must divide n is:
a)6
b)12
c)24
d)36
e)48
answer later
A lot of integer property questions can be solved using prime factorization.
For questions involving divisibility, divisors, factors and multiples, we can say:
If N is a factor by k, then k is "hiding" within the prime factorization of N
Consider these examples:
3 is a factor of 24, because 24 = (2)(2)(2)(3), and we can clearly see the 3 hiding in the prime factorization.
Likewise, 5 is a factor of 70 because 70 = (2)(5)(7)
And 8 is a factor of 112 because 112 = (2)(2)(2)(2)(7)
And 15 is a factor of 630 because 630 = (2)(3)(3)(5)(7)
---------------ASIDE #2--------------------------------------
IMPORTANT CONCEPT: The prime factorization of a perfect square will have an even number of each prime
For example: 400 is a perfect square.
400 = 2x2x2x2x5x5. Here, we have four 2's and two 5's
This should make sense, because the even number of primes allows us to split the primes into two EQUAL groups to demonstrate that the number is a square.
For example: 400 = 2x2x2x2x5x5 = (2x2x5)(2x2x5) = (2x2x5)²
Likewise, 576 is a perfect square.
576 = 2x2x2x2x2x2x3x3 = (2x2x2x3)(2x2x2x3) = (2x2x2x3)²
--------NOW ONTO THE QUESTION!------------------
Given: n² is divisible by 72 (in other words, there's a 72 hiding in the prime factorization of n²)
So, n² = (2)(2)(2)(3)(3)(?)(?)(?)(?)(?)... [the ?'s represent other possible primes in the prime factorization of n²]
Since we have an ODD number of 2's in the prime factorization, we can be certain that there is at least one more 2 in the prime factorization.
So, we know that n² = (2)(2)(2)(3)(3)(2)(?)(?)(?)(?)
So, while there MIGHT be tons of other values in the above prime factorization, we do know that there MUST BE at least four 2's and two 3's.
Now do some grouping to get: n² = [(2)(2)(3)(?)(?)...][(2)(2)(3)(?)(?)...]
From this we can see that n = (2)(2)(3)(?)(?)...
Question: What is the largest positive integer that must divide n?
(2)(2)(3) = 12.
So, the largest positive integer that must divide n is 12
Answer: B
Cheers,
Brent