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12 Days of Christmas GMAT Competition 2025 - Day 1: Each of the two

1 2 3 4 5 6 7

VidhiAgarwal1097

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15 Dec 2025, 11:43

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Smaller set range can be odd only when 2 is included( only even prime no.)
hence, Set 1: 2,3,5,7,11
Set 2 : 7,11,13,17,19 Range of other set is 19-7=12

Bunuel

Each of the two sets consists of five consecutive prime numbers, and the sets have exactly two primes in common. If the range of the set with the smaller sum of terms is odd, what is the range of the other set?

A. 3
B. 8
C. 10
D. 12
E. 17

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Let the first set of 5 consecutive prime numbers be: {2,3,5,7,11}
Let the second set be: {7,11,13,17,19}

With the 2 primes common being 7 and 11.

We can confirm that the range of the set with the smaller sum of terms is odd (11-2=9).

Range of the other set is 12 (19-7=12)

So, Answer is D

Bunuel

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D. 12
Set 1= 2,3,5,7,11
Set 2= 7,11,13,17,19
Range of set 1= 9 (odd)
Range of set 2= 19-7=12

Bunuel

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Bunuel

Given that only two values are the same and and in the first set the range is odd - this can only happen if the least value is 2 because all the rest of the prime numbers are odd. Odd-Even= Odd everything else would give an even range
So set 1 should be (2,3,5,7,11)
And set is (7,11,13,17,19)
Range of set 2 is thus 19-7 = 12
Ans D

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There are 2 sets of 5 consecutive prime numbers, with exactly 2 primes in common.

Range of set of smaller sum of terms is odd.

Some facts:
- Prime numbers are odd except 2
- Difference between 2 odd numbers are always even.
- Difference of an odd and even number can be odd.

So if the range is odd, it implies that one of the value is 2.

So, the smaller set would be (2,3,5,7,11)

and as 2 primes are common, the other set would be (7,11,13,17,19)

Hence the range is 19-7 = 12

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Bunuel

There are two sets: set 1 and set 2 which contains five consecutive prime numbers.

Set 1 = ( a,b,c,d,e)

Set 2= (d,e,f,g,h)

The common element among the two sets are two prime numbers.

Let’s consider set 1 as the lowest sum. The range of set 1 is ODD.

Range = highest value - lowest value.

Even - even = even

Odd - odd = even

Odd - even = odd. So, one of the term is even.

The only EVEN prime is 2.

Set 1 = ( 2,3,5,7,11)

Thus, Set 2 = (7,11,13,17,19)

The range of Set 2 = 19-7 = 12

Option D

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As we need range of smaller set of 5 consecutive prime numbers to be odd (last prime no. - first prime no.) , set must start with 2 as it is the only even prime number which can provide odd range (9-2 is 7 which is odd).
Hence smaller set = (2,3,5,7,11)
As Another set contains two prime numbers in common , set will be (7,11,13,17,19).
Hence, range of the other set = 19 - 7 = 12
Hence , option D.

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15 Dec 2025, 13:16

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2 consecutive sets - smaller set's sum is odd, that means all 5 need to be ODDS,

so the smaller set needs to be 3,5,7,11,13

and the other set has exactly 2 primes in common, it cant be something in between because it will give us more than two primes in common
Eg.5,7, 11,13,19 as there will 3 primes in common

So we will have the select only the 2nd last and last term

So the other set will be 11,13,17,19,23

so 23-11= 12

Option D

Bunuel

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The only way 2 sets of 5 consecutive numbers have 2 numbers common is for one set's largest 2 numbers to be the smallest 2 numbers in the second set. Hence the sets would look like:
A: {a, b, c, d, e}, B = {d, e, f, g, h}
Now, all prime numbers except 2 are odd. Hence, for the range of the set with the smaller sum of terms to be odd, it must contain at least 1 (and the only) even prime number that is 2. Hence A would look like:
A: {2, 3, 5, 7, 11} - Range is 11-2 = 9 (odd)
This means B would look like:
B: {7, 11, 13, 17, 19} - Range = 19-7 = 12

Bunuel

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Bunuel

Since the range of smaller set of primes is odd, basically says that the first term = 2,
since only even - odd =odd
so smaller set A= 2,3,5,7,9
larger set= 7,9,11,13,19
range= 19-7=12

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Set one:

2,3,5,7,9

Range: 9-2 = 7

Set Two:
7,9,11,13,17

Range: 17-7 = 10

Answer is 10 | Choice C

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Let smaller set A = a1,a2,a3,a4,a5
and set B = b1,b2,b3,b4,b5

Since exactly two primes are in common, a4,a5 = b1,b2

We know that only 2 is an even prime number and since the range of set A is odd, a1 must be even (odd - even = odd)

So, set A = 2,3,5,7,11
and set B = 7,11,13,17,19

Range of set B = 19-7 = 12

Bunuel

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Bunuel

For this, we will need to recall number properties. An odd number minus an odd number is always even. Since every prime number above 2 is odd, that means that the range of set A has an even number as its smallest value, which must be 2 (the only even prime number). This will make Set A's range an odd number.

Thus, Set A must be {2, 3, 5, 7, 11}.

The question says that there are two primes in common, and they're also sets of consecutive integers. The overlap of Set B can only be the last two numbers in Set A: 7 and 11.

From here, we can determine that Set B is: {7, 11, 13, 17, 19}. The range is then 19 - 7 = 12.

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1. First set can be 2, 3, 5, 7, 11 (This should be the set with the smaller sum)
2. The range of that set is 9
3. The second set should be 7, 11, 13, 17, 19
4. The range of that second set is 12
5. 7 and 11 share sets, so it fulfills the condition

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Set -1 (with smaller sum):
There are two conditions laid -
Five consecutive prime numbers
The range is odd i.e. (Biggest number - smallest number) in the set is odd

The second condition will be met only if the smallest prime number is 2
So, Set -1 of five consecutive prime numbers: 2, 3, 5, 7, 11

Set-2:
Again there are two conditions laid -
five consecutive prime numbers
both sets have exactly two primes in common

To meet the condition, the numbers must be last two of the Set-1, i.e. 7 & 11 only. Because, if we assume, 3 & 5 are matching numbers, since the sets are consecutive prime numbers, Set-2 will consists of 3, 5, 7, 11 & 13, i.e. total 4 matching numbers will be there
So, Set-2 will be: 7, 11, 13, 17, 19

So, the range of other set: 19 - 7= 12

Bunuel

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Facts: Set a & Set b, 5 consecutive prime, 2 prime in common, the lower sum set range is odd

I realize something is wrong with "the lower sum set range is odd", and guessing its probably unique form of set of prime, which likely related to the fact that 2 is the only even prime number

So I decided to just write this
Set A: 2,3,5,7,9
Set B: Since two number in common, so it must be 7,9,11,13,17

Range of set B is 17-7=10

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Prime numbers in order: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ........
Let the two sets be A and B.
Let set A be the one with the smaller sum of terms.

Since the two sets have 5 consecutive primes and have only two in common, the overlap must happen at the end of the first set (A) and start of the second set (B).

As the range for set A is odd, the first prime in set A must be 2
Why? 2 is the only even prime number. So, to get the range (Max - Min) as odd, the smallest number must be even.
Hence, set A is (2, 3, 5, 7, 11)
and set B must be (7, 11, 13, 17, 19)
So, the range of set B is 19 - 7 - 12

Hence, the answer is D

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Bunuel

Set A be set with smaller sum of terms, set B be set with larger sum of terms.

Some inferences:

The only way two set of consecutive primes can share two numbers are the two numbers that's at the end of set A.

If the range of set A is odd, that means one of the element high or low must be even and other must be odd.
This can be possible only if smallest element is 2

Therefore Set A = {2,3,5,7,11}
Set B = {7,11,13,17,19}

Range of Set B = 19-7 = 12

Correct Answer D

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Answer D)
list out the prime numbers
1, 2, 3, 5, 7, 11, 13, 17, 19
2-11 gives an odd range; therefore the range of the second set is 19-7=12

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Because the range of the set with smaller sum of term is odd, so the set should be {2,3,5,7,11)
and the other set will be {11,13,17,19,23} => range is 12

Bunuel