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Set A consists of all distinct prime numbers which are 2 more than a multiple of 3. If set B consists of distinct integers, is set B a subset of set A?

(1) Set B consist of two positive integers whose product is 10.

(2) The product of reciprocals of all elements in set B is a terminating decimal.

According to the stem set A = {2, 5, 11, 17, 23, 29, 41, ...}

(1) Set B consist of two positive integers whose product is 10.

10 can be broken into the product of two integers in two ways: 10 = 1*10 and 10 = 2*5. If set B = {1, 10}, then the answer to the question is NO but if set B = {2, 5}, then the answer to the question is YES. Not sufficient.

(2) The product of reciprocals of all elements in set B is a terminating decimal.

If set B = {4, 8}, then the answer to the question is NO but if set B = {2, 5}, then the answer to the question is YES. Not sufficient.

(1)+(2) Both possible sets from (1) meet the condition stated in the second statement: \(\frac{1}{1}*\frac{1}{10}=\frac{1}{10}=0.1= terminating \ decimal\) and \(\frac{1}{2}*\frac{1}{5}=\frac{1}{10}=0.1= terminating \ decimal\). Thus we still have two sets, which give two different answers to the question. Not sufficient.

How come you included 2 in the set A? Should the set A start with 5 instead of 2?

0 is a multiple of every integer, except 0 itself. Hence 0 + 2 = {a multiple of 3} + 2 = 2 = prime.

ZERO:

1. 0 is an integer.

2. 0 is an even integer. An even number is an integer that is "evenly divisible" by 2, i.e., divisible by 2 without a remainder and as zero is evenly divisible by 2 then it must be even.

3. 0 is neither positive nor negative integer (the only one of this kind).

4. 0 is divisible by EVERY integer except 0 itself.

Is there any easy way to tell if a fraction is a terminating decimal?

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 \(2*5^2\). 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.

These kind of questions teach you to very carefully approach every single GMAT problem. Zero is a multiple of every single integer - it is an often overlooked property of zero and it is the key to solve this question.

if a subset includes more than one copy of a member of the original set. Then is it still considered a subset? for example A= 1,2,3 and if B =1.2,2. Would B be a subset of A?

if a subset includes more than one copy of a member of the original set. Then is it still considered a subset? for example A= 1,2,3 and if B =1.2,2. Would B be a subset of A?

There are no two 2's in A, so no. A subset of a set can include only the elements of the initial sets.
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Set A consists of all distinct prime numbers which are 2 more than a multiple of 3. Can you please explain this bit clearly.

I can see the set as ( 2,5,11, etc) , here 3 and 7 are missing, but 7 is not a multiple of 3. Explain this point too.

Set A consists of all distinct prime numbers which are 2 more than a multiple of 3, so primes which are 2 more than a multiple of 3, so {a multiple of 3} + 2 = prime: {2, 5, 11, 17, 23, 29, 41, ...}

How does 3 or 7 satisfy {a multiple of 3} + 2 = prime?
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