M30-06

THEORY:

A reduced fraction \(\frac{a}{b}\) (meaning that the fraction is already in its simplest form, so reduced to its lowest term) can be expressed as a terminating decimal if and only if the denominator \(b\) 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 the denominator \(250\) equals \(2*5^3\). The fraction \(\frac{3}{30}\) is also a terminating decimal, as \(\frac{3}{30}=\frac{1}{10}\) and the denominator \(10=2*5\).

Note that if the denominator already consists of only 2s and/or 5s, 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 a terminating decimal.

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

Check Terminating and Recurring Decimals Problems in our Special Questions Directory.
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Official Solution:


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?

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

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

10 can be expressed as 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 the 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) satisfy the condition stated in the second statement: \( \frac{1}{1}*\frac{1}{10}=\frac{1}{10}=0.1\) (a terminating decimal) and \(\frac{1}{2}*\frac{1}{5}=\frac{1}{10}=0.1\) (a terminating decimal). Thus, we still have two sets, which give two different answers to the question. Not sufficient.


Answer: E
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Bunuel,

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. 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.

Check more here: https://gmatclub.com/forum/tips-and-hint ... l#p1371030
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Thanks!

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

I think this is a high-quality question and I agree with explanation.

I want to ask a 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?

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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Hi Bunuel,

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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Hi Bunuel,

I'm super confused as to how statement 2 works...the examples given in statement 2: (4,8) and (2,5) seem to be subsets of set A, yet (4,8) is said to be NO, and (2,5) is said to be YES. Can you please help me on this point? thanks.

Set A = {2, 5, 11, 17, 23, 29, 41, ...} (distinct prime numbers which are 2 more than a multiple of 3).

The question asks whether set B is a subset of set A.

If from (2) set B is {4, 8}, then the answer is NO: {4, 8} is NOT a subset of {2, 5, 11, 17, 23, 29, 41, ...}
If from (2) set B is {2, 5}, then the answer is YES: {2, 5} IS a subset of {2, 5, 11, 17, 23, 29, 41, ...}


Hope it's clear.
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chetan2u ScottTargetTestPrep AjiteshArun VeritasKarishma

Multiple of a number in my opinion does not take into account 0.

In general, if someone asks me , write down the multiples of 3.

I will write 3,6,9,12,15.......so on.

Do you think we should really take 0*3=0 as a case here??

Multiples can be negative, 0 or positive. Factors are positive only.
0 is a multiple of all numbers. So 0 is a multiple of 3 too.

Response:

As a matter of fact, 0 is a multiple of every integer, including itself.

An integer n is a multiple of an integer m if we can express n as n = m*k where k is an integer. So, in order to determine whether 0 is a multiple of 3, we should ask the following question: “Can we find an integer k such that 0 = 3*k?”. The answer is yes; if we let k = 0, we see that the equality holds. That’s why 0 is a multiple of 3 (and any other integer).
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Thank team for the good question. But I am of the opinion the answer is C - since Set (1,10) is not a subset to Set A . Hence, from statement 1 + 2 , only one set remains i.e (2,5)

Sets from Statement 1 - (1,10),(2,5)
Sets from Statement 2 - (2,5) (Multiples of 10)... infinite

The question asks: is set B a subset of set A? 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.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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