3. If 0 < x < y and x and y are consecutive perfect squares, what is the remainder when y is divided by x?

Notice that since x and y are consecutive perfect squares, then \(\sqrt{x}\) and \(\sqrt{y}\) are consecutive integers.

(1) Both x and y have 3 positive factors. This statement implies that \(x=(prime_1)^2\) and \(y=(prime_2)^2\). From above we have that \(\sqrt{x}=prime_1\) and \(\sqrt{y}=prime_2\) are consecutive integers. The only two consecutive integers which are primes are 2 and 3. Thus, \(x=(prime_1)^2=4\) and \(y=(prime_2)^2=9\). The remainder when 9 is divided by 4 is 1. Sufficient.

(2) Both \(\sqrt{x}\) and \(\sqrt{y}\) are prime numbers. The same here: \(\sqrt{x}=2\) and \(\sqrt{y}=3\). Sufficient.

Answer: D.
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5. If a, b, and c are integers and a < b < c, are a, b, and c consecutive integers?

Note that:
A. The factorial of a negative number is undefined.
B. 0!=1.
C. Only two factorials are odd: 0!=1 and 1!=1.
D. Factorial of a number which is prime is 2!=2.

(1) The median of {a!, b!, c!} is an odd number. This implies that b!=odd. Thus b is 0 or 1. But if b=0, then a is a negative number, so in this case a! is not defined. Therefore a=0 and b=1, so the set is {0!, 1!, c!}={1, 1, c!}. Now, if c=2, then the answer is YES but if c is any other number then the answer is NO. Not sufficient.

(2) c! is a prime number. This implies that c=2. Not sufficient.

(1)+(2) From above we have that a=0, b=1 and c=2, thus the answer to the question is YES. Sufficient.

Answer: C.
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8. Set A consist of 10 terms, each of which is a reciprocal of a prime number, is the median of the set less than 1/5?

(1) Reciprocal of the median is a prime number. If all the terms equal 1/2, then the median=1/2 and the answer is NO but if all the terms equal 1/7, then the median=1/7 and the answer is YES. Not sufficient.

(2) The product of any two terms of the set is a terminating decimal. This statement implies that the set must consists of 1/2 or/and 1/5. Thus the median could be 1/2, 1/5 or (1/5+1/2)/2=7/20. None of the possible values is less than 1/5. Sufficient.

Answer: B.

Theory:
Reduced fraction \(\frac{a}{b}\) (meaning that fraction is already reduced to its lowest term) can be expressed as terminating decimal if and only \(b\) (denominator) is of the form \(2^n5^m\), where \(m\) and \(n\) are non-negative integers. For example: \(\frac{7}{250}\) is a terminating decimal \(0.028\), as \(250\) (denominator) equals to \(2*5^3\). Fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and denominator \(10=2*5\).

Note that if denominator already has only 2-s and/or 5-s then it doesn't matter whether the fraction is reduced or not.

For example \(\frac{x}{2^n5^m}\), (where x, n and m are integers) will always be the terminating decimal.

We need reducing in case when we have the prime in denominator other then 2 or 5 to see whether it could be reduced. For example fraction \(\frac{6}{15}\) has 3 as prime in denominator and we need to know if it can be reduced.

Questions testing this concept:
does-the-decimal-equivalent-of-p-q-where-p-and-q-are-89566.html
any-decimal-that-has-only-a-finite-number-of-nonzero-digits-101964.html
if-a-b-c-d-and-e-are-integers-and-p-2-a3-b-and-q-2-c3-d5-e-is-p-q-a-terminating-decimal-125789.html
700-question-94641.html
is-r-s2-is-a-terminating-decimal-91360.html
pl-explain-89566.html
which-of-the-following-fractions-88937.html
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10. If N = 3^x*5^y, where x and y are positive integers, and N has 12 positive factors, what is the value of N?

\(N = 3^x*5^y\) has 12 positive factors means that (x+1)(y+1)=12=2*6=6*2=3*4=4*3. We can have the following cases:

\(N = 3^1*5^5\);
\(N = 3^5*5^1\);
\(N = 3^2*5^3\);
\(N = 3^3*5^2\).

(1) 9 is NOT a factor of N. This implies that the power of 3 is less than 2, thus N could only be \(3^1*5^5\). Sufficient.

(2) 125 is a factor of N. This implies that the power of 5 is more than or equal to 3, thus N could be \(3^1*5^5\) or \(3^2*5^3\). Not sufficient.

Answer: A.


THEORY.
Finding the Number of Factors of an Integer

First make prime factorization of an integer \(n=a^p*b^q*c^r\), where \(a\), \(b\), and \(c\) are prime factors of \(n\) and \(p\), \(q\), and \(r\) are their powers.

The number of factors of \(n\) will be expressed by the formula \((p+1)(q+1)(r+1)\). NOTE: this will include 1 and n itself.

Example: Finding the number of all factors of 450: \(450=2^1*3^2*5^2\)

Total number of factors of 450 including 1 and 450 itself is \((1+1)*(2+1)*(2+1)=2*3*3=18\) factors.
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Hi Bunuel,
Could you please clarify if 0 is a multiple of every number in GMAT land.
Thanks

0 is a multiple of every integer, except 0 itself.
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A small question, when the outcome of {a!, b!, c!} is an odd number ...is ...{1, 1, c!}, should we need to check what C is ? as whatever the C is it cant be consecutive integers.

Regards
Sanjiv

Hi Bunuel,

Could you please tell me (with sample numbers) how you arrived at the maximum value of [a] + [b] = -2 for any negative a and b?

i took the case of a=-2 and b=-1 and hence ab=2. In this case i am getting [a]+[b] = (-2) + (-1) = -3

Let me know if i am missing something.

Oh yes! It does make sense. I understand that the statement does not fall apart. But I was just interested in knowing if there is a case possible where we can get -2 as the maximum value.

Dear Bunuel,
First of all,i would like to thank you for all your sincere efforts that help innumerable gMAT takers like me.

In the explanation provided for statement 2,i did not quite understand how all the numbers of the set would have the same sign,if the smallest and largest terms have the same sign.The set contains>2 terms and from statement,we can comment only about 2 terms.Can you help me understand?

Thank you,
Fido