integers prep

Question

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

KarishmaB · Answer

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

jhudson · Answer

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.

jhudson · Answer

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?

KarishmaB · Answer

I think you first need to go through these two posts: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2010/12 ... ly-number/ https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/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)

jhudson · Answer

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?

jhudson · Answer

I HAVE IT!!!

jhudson · Answer

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?

KarishmaB · Answer

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.

jhudson · Answer

I am not paraphrasing. I made corrections to the question.

jhudson · Answer

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.

KarishmaB · Answer

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.

Vips0000 · Answer

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.

KarishmaB · Answer

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.

jhudson · Answer

Thanks for the response.

jhudson · Answer

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.

KarishmaB · Answer

I am not sure why you feel the answer must be -2 but this is why it is not: Ques3.jpg 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.

integers prep

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