Bunuel wrote:

Bunuel wrote:

Set T is a finite set of positive consecutive multiples of 14. How many of these integers are also multiples of 21?

(1) Set T consists of 30 integers.

(2) The smallest integer in set T is a multiple of 21.

Kudos for a correct solution.

VERITAS PREP OFFICIAL SOLUTION:Question Type: What Is the Value? This question asks: “How many numbers in Set T are multiples of 21?”

Given information in the question stem or diagram: Set T is a finite set of positive consecutive multiples of 14. It might be a good idea to think about the relationship between multiples of 14 and 21 before moving on. Multiples of 14 are multiples of 2 • 7, and multiples of 21 are multiples of 3 • 7. In order to be both, a number needs to be a multiple of 6 • 7, or 42. Indeed 42 is the smallest positive number that is a multiple of both 14 and 21. What you need to realize before moving to the statements is that every third multiple of 14 is also a multiple of 21.

Statement 1: Set T consists of 30 integers. If every third multiple of 14 is a multiple of 21, then there would be 10 multiples of 21 in Set T. A quick glance at Statement 2 indicates that it may be important to know if the first member of the set is a multiple of 21, which would make this statement alone not sufficient. However, since it is every third member of Set T that is a multiple of 21, and since 30 is a multiple of 3, there will be exactly 10 sets of 3, and each of those will have one multiple of 21. It does not matter if the multiple of 21 is the first number in each group of 3 or the last. This is because 30 is an exact multiple of 3. It the set contained 31 integers, then it would be important to know if the multiple of 21 was first, but in this case you do not need that information. This statement is sufficient alone. The answer is either answer A or D.

Statement 2: The smallest integer in Set T is a multiple of 21. This statement is not sufficient alone since it does not indicate the number of members of the set.

The correct answer is A. Note: This question is a very tricky example of the “Why Are You Here?—Temptation” construct. A good test-taker will very carefully analyze this second statement and prove whether it matters in relation to the first. In this case, it does not matter, so the answer is A, but in other cases it certainly will, and the answer would be C. The key here is that you take the time to make sure that the second statement does or does not matter.

Adding to this explanation..

Why is it that when the Number of elements(say n) is a multiple of some number(say x), there would be exactly n/x numbers that are multiples of x?This is pure visualization...

Considering this example,

30 is a multiple of many factors other than 3(Ex 2,5,6,10..)

But we will first consider 3, and that should explain it all..

Now,

Imagine some consecutive numbers written in front of you..

\(a_1 a_2 a_3 a_4 a_5 a_6 a_7 ... a_{28} a_{29} a_{30}\)

because there are 30 numbers..we can..starting from the left..group them into groups of three numbers and leave nothing behind..thus our groups would look like..

\((a_1 a_2 a_3), (a_4 a_5 a_6), ....(a_{28} a_{29} a_{30})\)

we know that out of any n consecutive integers..only and only 1 number is necessarily a multiple of n..

Now look the groups we just made..

We will have exactly 10 groups here of 3 consecutive numbers each..and exactly 1 of the numbers in every group will be a multiple of 3.

Thus, there will be exactly 10 numbers that are multiples of 3If \(a_1\) is a multiple of 3..then the numbers will be

\(a_1 a_4 a_7.....a_{28}\)

i.e. the first number of every group we just made.

if \(a_2\) is multiple of 3..then every second number of the groups will be a multiple of 3..and every third if \(a_3\) is a multiple of 3.

In any case, we will have exactly 10 numbers that are multiples of 3 in this seriesWhat if there were 31 numbers?If you start grouping the numbers in the same manner..you will have \(a_{31}\) left in the last as it can't be grouped.

This is where and how the discrepancy can take place...

If \(a_1\) is a multiple of 3..then the multiples of 3 will be..

\(a_1 a_4 a_7... a_{28} a_{31}\)

i.e. every first number of the groups. Here we have 11 numbers that are multiples of 3.

If \(a_2\) is a multiple of 3..then the multiples of 3 will be..

\(a_2 a_5 a_8... a_{29}\)

i.e. every second number of the groups. Here we have 10 numbers that are multiples of 3. Why?

Because in the last group starting with \(a_{31}\)..the second and the third numbers are not there!

This is why we would need to check case by case as 31 is not a multiple of 3.

Wanna check for multiples of 5? divide the 30 numbers into 6 groups of 5 numbers each..the same would apply..

I hope you enjoyed

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