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

1 2 3 4 5 6 7

redandme21

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In any set of consecutive prime numbers the range is even except if the set has the number 2.
First set: 2, 3, 5, 7, 11
Second set with two primes in common: 7, 11, 13, 17, 19

Range of the second set is 19-7=12

IMO D

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The range of the set with the smaller sum of terms is odd, so the only possible set must contain "2". As this is the smallest prime number, the set must be {2, 3, 5, 7, 11}. The other set has exactly two primes in common, so this set must be {7, 11, 13, 17, 19} whose range is 19-7=12.

Answer D

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Smaller sum is odd => 2 belong there
Smaller Sum Set = {2,3,5,7,11}
2 elements common i.e. 7,11 common
Larger Sum Set = {7,11,13,17,19}
Range of Larger Sum Set = 19 - 7 = 12

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15 Dec 2025, 20:38

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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

Consecutive terms - 2,3,5 ,7 11,13....

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

RANGE OF A = 9
RANGE OF B = 12

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The only even prime number is 2. The difference between two odd numbers is an even number. If the range is odd, the first number must be 2.
setA: 2 3 5 7 11
setB must contain 7 and 11: 7 11 13 17 19

range of setB is 12 (19-7)

The answer is D

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For five consecutive primes, the only way the range is odd is when the set includes 2

So the smaller set must be: 2,3,5,7,11

Now the two sets have exactly two primes in common.
List of consecutive prime sets that share exactly two primes with this set:
And larger set with common primes: 7,11,13,17,19
So the range is : 12

Correct option: D

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If the range of the set with the smaller sum of terms is odd the smallest number must be even and the biggest number must be odd. 2 is the only even prime number.
The set is 2, 3, 5, 7, 11
The other set must contain 7 and 11: 7, 11, 13, 17, 19 and its range is 19-7=12

The correct answer is D

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A 3
The number needs to be small and odd.

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If two blocks of size n overlap by K elements , shift = n-k , here (5-2 =3 ).
Set 1 = {p1,p2,p3,p4,p5}
Set 2 = { p4,p5,p6,p7,p8}
p4 and p5 - exactly 2 in common
Set 2 replaces the first 3 primes of set 1 with next 3 larger primes.
Set 2 has larger sum . set 1 has smaller sum.
Range of set with smaller terms is odd .
odd - odd = even
odd- even = odd
Set must include the prime number 2
First Set = {2,3,5,7,11} , Range 11-2 =9
Second set = {7,11,13,17,19} , Range 19-7 = 12 (Final Answer)

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The key to the answer lies in this statement -range of a set of prime numbers is odd. If the difference of two numbers are odd, this means one number is odd, other is even.
Since 2 is the only even prime number, now we know that 2 is included in set 1 of smaller numbers.
Given that the set consists of 5 consecutive prime numbers, we get the smaller set as 2 3 5 7 11
Since there is an overlap of two numbers. we get the second set as 7 11 13 17 19.
The range of this set is 19 - 7 = 12.

Bunuel

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So for any subtraction to be odd we need one number even and one odd.
Range is given by (largest no in set- smallest no in set)

so only prime number which is also even is 2

so first set starts with 2.
first set set - 5 consecutive prime no is ( 2,3,5,7,11)
second set has only one no common with first set and it is also a set of 5 consecutive no
so we get second set as (11,13,17,19,23)

range on second set is (23-11)=12

answer is Option D:12

Bunuel

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Bunuel

Given,
Set A : {p1, p2, p3, p4, p5}
Since, Set B has 2 primes in common with Set A ->
Set B : {p4, p5, p6, p7, p8}

Range of Set A is Odd i.e., (p5 - p1) is odd

All prime numbers are Odd except for 2. And (Odd - Odd) is always even and (Odd - Even) is always odd.

Based on this, we can conclude that the first term of Set A must be 2.

So,
Set A : {2, 3, 5, 7, 11}
Set B : {7, 11, 13, 15, 19}

Range of Set B = (19 - 7) = 12

Therefore, answer D) 12

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Each set has 5 consecutive prime numbers and they share exactly 2 numbers.
Its given that smaller set range is Odd (Range is the difference between highest and smallest number in a set)
If a set only contains odd prime numbers the range is Even, since the smaller set has an Odd range, it must include 2

Hence:
Set A - {2,3,5,7,11}
(Range 11 - 2 = 9 (odd))

Set B - {7,11,13,17,19}
Hence, the range is 19-7 = 12
Option D

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If the range of the set of 5 prime consecutive numbers is odd, it is fair to assume one number to be even & the other to be odd. Since there is only one even prime number which is 2 & it is the least big prime number, the set that it would be a part of to satisfy the condition of 5 consecutive prime numbers would be (2,3,5,7,11).

If there are 5 other prime numbers in another set that have only two elements in common with the above set & are consecutive prime numbers then that set would contain the largest two prime numbers which in this case is 7 & 11. Hence the set would be (7,11,13,17,19). The difference between 19 & 7 = 12 which is the range of this set

Hence the answer to this question is option (D) 12

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Given the range of first set is odd i.e one of the number in set is 2 as 2 is the only even digit prime number. therefore the set will be (2,3,5,7,11).

Given that only two numbers are common in the both sets and both sets contain consecutive primes so only 7,11 will be the common primes. the next set will be (7,11,13,17,19)

The range will be = 19-7 =12.

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We know that, 2 is the only even prime number and difference between odd primes are even.

1st Set
Therefore, for a set of 5 consecutive primes to have an odd range, it must include 2. Hence, the only such block of 5 consecutive primes is - (2,3,5,7,11)

2nd Set
It is given that in the sets has 2 prime numbers in common, which means the 2nd set must have 7 and 11, and the complete set of 5 consecutive prime number is - (7,11,13,17,19)

Range of this set is = 19 - 7 = 12.
Hence the correct answer is D. 12

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For the range of the set with the smaller sum to be odd on of the digits must be even which implies that this set contains the number 2 (only even prime number)
=> s1 = [2, 3, 5, 7, 11] and s2 = [7, 11, 13, 17, 19] (as there are two prime numbers common)
=> range of s2 = 19 - 7 = 12

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As the range of the smaller set is odd... We can clearly assume it contains the lowest term as even prime number which is 2 and so, the smaller set will be {2,3,5,7,11} .. As the larger set contains exactly two prime numbers in common with the smaller set , we can safely assume those two consecutive numbers will be 7,11.. So, the larger set will be {7,11,13,17,19}. Sp, the range of the larger set is 19-7 which is 12. So, the answer is choice: D

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since odd - even = odd, we can clearly see that the first set starts with even
The only even prime is 2 so the set 1 is [2,3,5,7,11] and set 2 is [7,11,13,17,19]
The question asked for range of set 2 = 19-7 = 12
Option D

Bunuel

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The prime numbers is 1,2,3,5,7,11,13,17,19,23,25
If smallest set is A then {1,2,3,5,7}
Range = 7-1 = 6 which is even
So consider the set A as
{2,3,5,7,11}
Range = 11- 2 = 9 is odd
Then set B must be {11,13,17,19,23} since one number is common
Thus range = 23-11 = 12