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integers prep [#permalink] New post 08 Jul 2013, 17:14
If 60xn is the square of an integer, what is the least possible value that n could have?
a: 6
b: 9
c: 12
D: 15
E: 60
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Re: integers prep [#permalink] New post 09 Jul 2013, 00:57
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jhudson wrote:
If 60xn is the square of an integer, what is the least possible value that n could have?
a: 6
b: 9
c: 12
D: 15
E: 60


Recall that squares of integers have even powers of primes i.e. 9 = 3^2, 16 = 2^4, 36 = 2^2*3^2 etc

60 = 2^2*3*5

Notice that to make a perfect square, we need at least a 3 and a 5.

60n = 2^2*3^2*5^2 (all even powers)

So n must be at least 3*5 = 15
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Re: integers prep [#permalink] New post 09 Jul 2013, 07:12
VeritasPrepKarishma wrote:
jhudson wrote:
If 60xn is the square of an integer, what is the least possible value that n could have?
a: 6
b: 9
c: 12
D: 15
E: 60


Recall that squares of integers have even powers of primes i.e. 9 = 3^2, 16 = 2^4, 36 = 2^2*3^2 etc

60 = 2^2*3*5

Notice that to make a perfect square, we need at least a 3 and a 5.

60n = 2^2*3^2*5^2 (all even powers)

So n must be at least 3*5 = 15





So since 60 is multiplied by an integer, that allows us to make 3 and 5 a multiple of the square to make that number a perfect square. What I don't see is how we take only the 3 and 5 to determine the lowest possible value is 15. Why not 2 and 3? Sorry if this is confusing but I'm just getting back into these rules and things aren't clicking like they use too.
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Re: integers prep [#permalink] New post 09 Jul 2013, 07:17
jhudson wrote:
VeritasPrepKarishma wrote:
jhudson wrote:
If 60xn is the square of an integer, what is the least possible value that n could have?
a: 6
b: 9
c: 12
D: 15
E: 60


Recall that squares of integers have even powers of primes i.e. 9 = 3^2, 16 = 2^4, 36 = 2^2*3^2 etc

60 = 2^2*3*5

Notice that to make a perfect square, we need at least a 3 and a 5.

60n = 2^2*3^2*5^2 (all even powers)

So n must be at least 3*5 = 15





So since 60 is multiplied by an integer, that allows us to make 3 and 5 a multiple of the square to make that number a perfect square. What I don't see is how we take only the 3 and 5 to determine the lowest possible value is 15. Why not 2 and 3? Sorry if this is confusing but I'm just getting back into these rules and things aren't clicking like they use too.



Is this simply solving the equation for N. i.e. 2^2*3*5n=2^2*3^2*5^2 which equals n=3*5 based on cancellation?
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Re: integers prep [#permalink] New post 09 Jul 2013, 21:31
Expert's post
jhudson wrote:
jhudson wrote:
VeritasPrepKarishma wrote:
Recall that squares of integers have even powers of primes i.e. 9 = 3^2, 16 = 2^4, 36 = 2^2*3^2 etc

60 = 2^2*3*5

Notice that to make a perfect square, we need at least a 3 and a 5.

60n = 2^2*3^2*5^2 (all even powers)

So n must be at least 3*5 = 15





So since 60 is multiplied by an integer, that allows us to make 3 and 5 a multiple of the square to make that number a perfect square. What I don't see is how we take only the 3 and 5 to determine the lowest possible value is 15. Why not 2 and 3? Sorry if this is confusing but I'm just getting back into these rules and things aren't clicking like they use too.



Is this simply solving the equation for N. i.e. 2^2*3*5n=2^2*3^2*5^2 which equals n=3*5 based on cancellation?



I think you first need to go through these two posts:

http://www.veritasprep.com/blog/2010/12 ... ly-number/
http://www.veritasprep.com/blog/2010/12 ... t-squares/

If 60n is to be a perfect square (smallest since n must have the least possible value), all its prime factors must have an even power.
60 = 2^2 * 3* 5
2 already has an even power i.e. 2.
3 has an odd power 1 so we need another 3 to make an even power.
5 has an odd power 1 too so we need a 5 too so give 5 a power of 2.

2^2*3^2*5^2 will be a perfect square. It will be the smallest perfect square closest to 60 (so that n has the least possible value)
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Re: integers prep [#permalink] New post 11 Jul 2013, 14:36
Thanks for all your help, I really appreciate it. I have two more questions.

1. Why am I to assume 60N= a perfect square from the original question?

2. I did not understand the passage below:


"If we want to find out its even factors, we multiply each of the odd factors by 2 or 2^2. We can take 2 in two ways (one 2 or two 2s). We cannot take no 2 because that just leaves us with an odd factor. So we get 3 × 2 = 6 even factors. A perfect square always has even number of even factors."

Where is the 2^2 or 4 coming from? Are we just taking the exponent from the "2" and multiplying that against the odd factors?
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Re: integers prep [#permalink] New post 11 Jul 2013, 15:21
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Re: integers prep [#permalink] New post 15 Jul 2013, 15:41
6PQ is a 3-digit number (P and Q are digits that is divisible by 4. When 6PQ is divided by 5, the remainder is 2. How many different values can P take?
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Re: integers prep [#permalink] New post 15 Jul 2013, 21:11
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jhudson wrote:
6PQ is a 3-digit number (P and Q are digits that is divisible by 4. When 6PQ is divided by 5, the remainder is 2. How many different values can P take?


When you write down the question, do not paraphrase. Write it exactly as given. What does "(P and Q are digits that is divisible by 4." mean? I am assuming it means that the two digit number PQ is divisible by 4. Each P and Q cannot be divisible by 4 because then Q can take only one of 3 values (0, 4, 8) but in no case will it give 2 remainder when divided by 5.
To get a remainder of 2, the last digit of the number i.e. Q must be either 2 or 7 (2 more than a multiple of 5). If Q is 7, PQ is odd and hence is not divisible by 4. Hence Q must be 2.
If Q is 2 and PQ is divisible by 4, P must be 1 or 3 or 5 or 7 or 9 i.e. it can take 5 values.
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Re: integers prep [#permalink] New post 16 Jul 2013, 17:11
jhudson wrote:
6PQ is a 3-digit number (P and Q are digits) that is divisible by 4. When 6PQ is divided by 5, the remainder is 2. How many different values can P take?


I am not paraphrasing. I made corrections to the question.
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Re: integers prep [#permalink] New post 16 Jul 2013, 18:07
I have the answer to the question. I do not understand how to determine all 5 values or know you have found all 2-digit numbers that make of the possible options.
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Re: integers prep [#permalink] New post 16 Jul 2013, 19:21
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jhudson wrote:
I have the answer to the question. I do not understand how to determine all 5 values or know you have found all 2-digit numbers that make of the possible options.


You missed the closing bracket and that created confusion.

As for the explanation, I have given it above. Nevertheless, if you are looking for a more detailed solution, here it is:

Think of the divisibility rule of 5. Every number which ends with 5 or 0 is divisible by 5. What happens if a number ends with 1 or 6? The number previous to it which ends with 0 or 5 will be divisible by 5 so remainder will be 1. e.g. 151126 will give remainder 1 when divided by 5 because 151125 is divisible by 5.

To get a remainder of 2 on division by 5, the last digit of the number i.e. Q must be either 2 or 7 (2 more than a multiple of 5). Notice that Q cannot be anything other than 2 or 7 since the remainder will be different in that case. e.g. if Q = 4, remainder when you divide the number by 5 will be 4.


If Q is 7, 6PQ is 6P7 i.e. an odd number. Hence it is certainly not divisible by 4. Hence Q must be 2.

If Q is 2 and 6P2 is divisible by 4, P must be 1 or 3 or 5 or 7 or 9 i.e. it can take 5 values. Just think of the divisibility rule of 4. For a number to be divisible by 4, the last two digits must be divisible by 4. Hence if P2 is divisible by 4, so is 6P2.

What values can P take such that P2 is divisible by 4?
Can P be 0? No.
Can P be 1? Yes, 12 is divisible by 4.
Can P be 2? No, 22 is not divisible by 4.
Can P be 3? Yes, 32 is divisible by 4.

Notice the pattern. For every odd value of P, P2 is divisible by 4. There are 5 odd values so there will be 5 solutions. This is something you should quickly figure in your head.
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Re: integers prep [#permalink] New post 16 Jul 2013, 19:52
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Alright, I want to play a spoil sport here... Basically question is not correct or the answers...

There is no restriction on values of n, therefore value of n could be a fraction 1/60 or a 0 and you get the least value of n, while,satisfying condition of 60n being a perfect square.

With the given answers question must have specified- n is a nonzero integer and then the solutions given above work.
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Re: integers prep [#permalink] New post 16 Jul 2013, 20:21
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Vips0000 wrote:
Alright, I want to play a spoil sport here... Basically question is not correct or the answers...

There is no restriction on values of n, therefore value of n could be a fraction 1/60 or a 0 and you get the least value of n, while,satisfying condition of 60n being a perfect square.

With the given answers question must have specified, n is a nonzero integer and then the solutions given above work.


An actual GMAT question will not have any ambiguities. Meanwhile, here the intent of the question is clear from the options. But yeah, you are right.
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Re: integers prep [#permalink] New post 21 Jul 2013, 11:37
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Re: integers prep [#permalink] New post 23 Jul 2013, 15:56
x<=-1/2, what is the greatest integer value of x?
a. -2
b. -1
c. 0
d. 1
e. 2

The answer is -1. Please explain why. Why not a. -2.
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Re: integers prep [#permalink] New post 23 Jul 2013, 19:55
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jhudson wrote:
x<=-1/2, what is the greatest integer value of x?
a. -2
b. -1
c. 0
d. 1
e. 2

The answer is -1. Please explain why. Why not a. -2.


I am not sure why you feel the answer must be -2 but this is why it is not:

Attachment:
Ques3.jpg
Ques3.jpg [ 3.43 KiB | Viewed 375 times ]


If x is less than -1/2, x lies in the region shown by the green arrow. Which is the greatest integer in that region? -1. So greatest integer value that x can take is -1.
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Re: integers prep   [#permalink] 23 Jul 2013, 19:55
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