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Range Problem.,...

by Viren1808 » Sat Nov 12, 2011 9:54 am

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Set X consists of prime numbers {3, 11, 7, K, 17, 19}. If integer Y represents the product of all elements in set X and if 11Y is an even number, what is the range of set X?

(A) 14 (B) 16 (C) 17 (D) 20 (E) 26
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by GmatMathPro » Sat Nov 12, 2011 10:13 am
Viren1808 wrote:Set X consists of prime numbers {3, 11, 7, K, 17, 19}. If integer Y represents the product of all elements in set X and if 11Y is an even number, what is the range of set X?

(A) 14 (B) 16 (C) 17 (D) 20 (E) 26
Y is the product of all of the elements and 11Y is even. 11 is odd, so for 11Y to be even, Y must be even. A product of a set of integers is even if and only if at least one of the integers is even. All of the numbers in the set thus far are odd, so if the product is even, K must be even. The only even prime number is 2, so K=2. The range is equal to the largest value minus the smallest value, so range=19-2=17

Ans: c. 17
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by neelgandham » Sat Nov 12, 2011 10:26 am
Viren1808 wrote:Set X consists of prime numbers {3, 11, 7, K, 17, 19}. If integer Y represents the product of all elements in set X and if 11Y is an even number, what is the range of set X?

(A) 14 (B) 16 (C) 17 (D) 20 (E) 26
Y = Product of all the elements in set X = 11*3*7*K*17*19
11y = 11*11*3*7*K*17*19 is an even number. For the product to be even the number K should be an even number and a prime. The only even prime in the number set is 2. So, the value of K is 2.Range of the set is equal to the difference between the highest and the least number.

Range = 19-2 = 17 Answer C
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by JohnRapp » Sat Nov 12, 2011 1:07 pm
I worked backwards on this one. The question said 11 x product of the set = even. in other words, odd (11) x even = even. Thus, the product of the set must be even. The question tells us that the set consists of only prime numbers. We know that the only even and prime number is 2. So K must be 2.

Greatest number in the set = 19.

19 - 2 = [spoiler]17 (C)[/spoiler]
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by masoom j negi » Fri Dec 21, 2018 8:58 pm
11Y will be even if Y is even.
Y = product of all elements in the set.
There is no even no in the set X. So, 'k' should be even to make y even.
There is only one even prime no i.e. 2. So, k = 2
The set X becomes {2, 3, 7, 11, 17, 19}.
Range of the set = highest element - lowest element = 19 - 2 = 17
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by Scott@TargetTestPrep » Sat Mar 02, 2019 8:05 am
Viren1808 wrote:Set X consists of prime numbers {3, 11, 7, K, 17, 19}. If integer Y represents the product of all elements in set X and if 11Y is an even number, what is the range of set X?

(A) 14 (B) 16 (C) 17 (D) 20 (E) 26
Since 11Y is even, Y is even and hence K must be even also. Since 2 is the only even prime number, K is 2. Thus, the range of set X is 19 - 2 = 17.

Answer: C

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by [email protected] » Sat Mar 02, 2019 1:16 pm
Hi All,

We're told that Set X consists of PRIME numbers {3, 11, 7, K, 17, 19}, the integer Y represents the PRODUCT of all elements in Set X and 11Y is an EVEN number. We're asked for the RANGE of Set X. This question is based on several Number Property rules - and doesn't actually require much math to solve.

To start, when dealing with a RANGE, it helps to put the existing elements in order from least to greatest: {3, 7, 11, 17, 19.... and K}. It's important to note that we do NOT know what K is - we know that it's PRIME, so it could be as small as 2 or as large as whatever the largest possible prime is...

Notice how the 5 numbers that we are given are ALL ODD numbers; thus, the product of those 5 ODDS would be a larger ODD number. We're told that Y is the PRODUCT of the 6 numbers.... so Y = (big-odd-number)(K). However, we're told that 11Y = EVEN; that can only happen when Y is an EVEN number.

Thus, if Y = EVEN = (big-odd-number)(K) then K must also be EVEN. The only even prime number is 2, which makes Set X = {2, 3, 7, 11, 17, 19} with a range of 19 - 2 = 17

Final Answer: C

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