12 Days of Christmas GMAT Competition 2025 - Day 3: A computer program

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Bunuel

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17 Dec 2025, 10:00

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Bunuel

GMAT Club Official Explanation:

A generated number not having a factor p such that 1 < p < the generated number means that the program generates only prime numbers, because a prime number does not have any factor greater than 1 and less than itself; it has only two factors: 1 and itself.

So, we need to find all possible different remainders that can occur when a prime with two or more digits is divided by 18:

prime = 18q + remainder

Notice here that the remainder must be less than 18, because dividing by 18 always produces a remainder from 0 to 17.

If 18 and the remainder share any factor greater than 1, then we could factor that out from 18q + remainder, and the expression would no longer represent a two or more digits prime number number. For example, if the remainder is 2, then we get 18q + 2 = 2(9q + 1), which cannot be a two or more digits prime number.

Thus, the remainder can only be a number less than 18 that shares no divisor greater than 1 with 18. Therefore, the remainder could be 1, 5, 7, 11, 13, or 17.

For example:

prime = 19 gives remainder 1
prime = 23 gives remainder 5
prime = 43 gives remainder 7
prime = 29 gives remainder 11
prime = 31 gives remainder 13
prime = 53 gives remainder 17

Answer: D.

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17 Dec 2025, 09:05

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When any number is divided by 18, the possible reminders are 0 through 17.
However, a prime number greater than 10 cannot be divisible by 2 or 3, so its reminder when divided by 18 cannot be multiple of 2 or 3
The numbers between 0 and 17 are:
1,5,7,11,13,17

All of these actually occur as reminders for primes greater than 10
So the number of distinct reminders possible is 6

Correct answer: D

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17 Dec 2025, 09:06

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p is a prime number
it can only be an odd number since p > 10
so all possible odd remainders will be the answer
1 3 5 7 9 11 13 15 17
answer is 9

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17 Dec 2025, 09:12

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No factor p such that 1 < p < generated number, means the number is prime

Possible remainders when 18 divides a number 0<=r<18

Given prime greater than 10 are considered, and 18 is even, the remainder cannot be even

So 0,2,4,6,8,10,12,14,16 are out
1,3,5,7,9,11,13,15,17 remains

Now if something has a remainder of 3 or 9 or 15 then they will be of the form 18k+3 => 3(6k+1) or 18k+9 => 9(2k+1) or 18K + 15 => 3(6k+5). Now this makes the number have 3 and 9 as factors, so these are out too.

Remaining 1,5,7,11,13,17

So 6 possible remainders

Answer D

Bunuel

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17 Dec 2025, 09:14

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Bunuel

[Deconstructing the Question
The problem defines the generated numbers as integers greater than 10 that have "no factor $p$ such that $1 < p < \text{number}$".
This is the definition of a Prime Number.
So, we are looking for the possible remainders when a prime number $n > 10$ is divided by 18.

Step 1: Properties of Primes > 10
Let the prime be $n$. Since $n > 10$, $n$ cannot be even (otherwise it's divisible by 2).
Since $n > 10$, $n$ cannot be a multiple of 3 (otherwise it's divisible by 3).
The divisor is $18 = 2 \times 3^2$.
Since $n$ shares no factors with 18, $n$ and 18 are coprime ($gcd(n, 18) = 1$).

Step 2: Analyze Remainders
If $n$ is coprime to 18, the remainder $r$ must also be coprime to 18.
The possible remainders modulo 18 are integers from 0 to 17. We must exclude all even numbers and all multiples of 3.
Let's list the candidates:
1 (Coprime) * 2, 3, 4 (Share factors with 18)
5 (Coprime) * 6 (Shares factors)
7 (Coprime) * 8, 9, 10 (Share factors)
11 (Coprime) * 12 (Shares factors)
13 (Coprime) * 14, 15, 16 (Share factors)
17 (Coprime) The set of possible distinct remainders is {1, 5, 7, 11, 13, 17}.

Step 3: Verification
Do primes exist for these remainders?
$19 \equiv 1 \pmod{18}$
$23 \equiv 5 \pmod{18}$
$43 \equiv 7 \pmod{18}$
$11 \equiv 11 \pmod{18}$
$13 \equiv 13 \pmod{18}$
$17 \equiv 17 \pmod{18}$
All are valid primes > 10.

Total count: 6.

Answer: D

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17 Dec 2025, 09:15

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9. Number of remainders for divisor 18 is 0 to 17 remainder. Numbers starts at greater than 10. Must repeat remainders after 18. 18 has remainder of zero

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17 Dec 2025, 09:19

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The program essentially throws out prime number greater than 10.
These numbers are all odd, therefore on division with 18, they will leave odd remainder.
Therefore possible remainders = {1,3,5,7,...17} = 9 in number
However, since they are all prime, they cannot leave multiple of 3 as remainder,
since if say 3 is a remainder of n, then n = 18q+3 which is a multiple of 3 and not a prime.
So removing 3,9 and 15 from list of 9 remainders, we are left with 6 remainders.

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17 Dec 2025, 09:23

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Any number greater than 10 has no factor p means it should be a prime number.
Now prime numbers from 11 onwards are:
11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61,....
When divided by 18 these yields remainder of:
11, 13, 17, 1, 5, 11, 13, 1, 5, 7, 11, 17,....
Seems only six of them are repeating.

OR

any number beyond 10 will have maximum of remainders: 1, 2, 3, 4, ....., 17.
Since the numbers from 1-17 will have multiples of 2 and 3 we can remove them because they cannot be remainders of prime.
So we are left with: 1, 5, 7, 11, 13, 17.

Hence: D

Bunuel

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17 Dec 2025, 09:23

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So we know that the computer program is generating prime numbers greater than 10.
Since the number is prime number and the program divides it by 18 (18=2*3^2), the remainder must be less than 18 and not divisible by 2 or 3.
=> The remainder: 1,5,7,11,13,17

Answer: D

Bunuel

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Answer D
The definitions of the number generated (let's call it n) tells us that each number generated n is a prime number (since it does not have any factor ≠ 1 or ≠ n), it also tells us that n > 10, so n could be any prime number greater than 10
n=x*18 + remainder=x(2*3^2) + remainder

if n/18 gave a remainder that is an even number this would mean that n is even, but it cannot be, since n is a prime number. (eg 20/18 -> remainder 2, 34/18 -> remainder 16)
if n/18 gave a remainder that is a multiple of 3 this would mean that n is a multiple of 3, but it cannot be.

Since n is a positive integer, the remainder must be an integer >= 1 and <18.
If from all those numbers we eliminate all even number and all multiples of 3, there are 6 numbers left: 1, 5, 7, 11, 13, 17

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17 Dec 2025, 09:24

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If the number is greater than 10 with the condition that p is not a factor of 1<p<generated number.
considering that p is given as a variable and could be any lower number that could divide the generated number unless the generated number is prime. In that case there would be no number greater than 1 but smaller than the generated number that could divide it.

Hence the generated number is a prime number greater than 10.

Now the prime greater than 10 would be 11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and so on.

Dividing these with 18 would give remainders as 11, 13, 17, 1, 5, 11, 13, 1, 5, 7 and so on. The list includes 1 and all the prime numbers smaller than 18 except 3.

Considering the fact that each prime number greater than 3 is of the form (multiple of 6 + 1) OR (multiple of 6 - 1), 3 cannot be the remainder as 18 is also a multiple of 6.

Hence, the list highlighted above is complete and there are only 6 such remainders. Answer is D.

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17 Dec 2025, 09:26

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A computer program randomly generates positive integers greater than 10 that have no factor p such that 1 < p < the generated number. Each time a number is generated, the program divides it by 18 and records the remainder. How many distinct remainders could the program possibly record?
A. 3
B. 4
C. 5
D. 6
E. 9

the condition of have no factor 1<p< the generated number is possible only when generated number is prime
possible numbers generated
11,13,17,19,23,27,29,31,37 so on...
remainders distinct possible when 18 is divided to generated numbers ( 11,13,17,19,23,27,29,31,37)

11,13,17,1,5,7

6 possible distinct remainders

OPTION D , 6 is correct

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17 Dec 2025, 09:30

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IMO D

The program generates positive integers greater than 10 with no factors other than 1 and itself that means we are looking for prime numbers greater than 10.

We are asked how many distinct remainders while dividing from 18 such primes can produce.

prime number>10 are 11, 13, 17, 19, 23, ... so on basicalaly.

we also need to know that since 18 has factors 2,3,3. Remainders cannot share factor with it, so we are only left with all other remainders that are prime, <18 and not 2,3.

those are 1,5,7,11,13,17. After this it will repeat, so 6 is the correct answer.

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17 Dec 2025, 09:31

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A number (N) which has no factor p such that 1<p<N will always be a prime number.
Given N>10. All prime numbers will always be odd numbers.
If we divide N by 18 then remainder (r) can only be 0 to 17.
N=18k + r, here N is odd, 18k is even. N can be odd only when r is also odd. (Odd=Even+Odd)
r can be 1,3,5,7,9,11,13,15,17.
Because N is a prime number so 18k and r should also be coprime to each other.
Hence, r must be 1,5,7,11,13,17
Answer is 6 (D)

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17 Dec 2025, 09:33

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We know:
-- that remainder will have to be lesser than 18
-- P is a prime number

p = 18x + r (x is an integer and r is the remainder)
Now r cannot be a multiple of 2 because if it is a multiple of 2 then:
p = 18x + 2R -> 2(9x + R) -> this will mean that p can be divided by 2 which should not be the case as p = 2 not possible (p>10) and p is a prime number.

r can also not be a multiple of 3 as if it is a multiple of 3:
p = 18x + 3R -> 3(6x + R) -> this will mean that p is divisible by 3 but that is not possible as p is prime and more than 10.

So possible values of r which are less than 18, odd, not multiples of 3 -> 1, 5, 7, 11, 13, 17

So, 6 values. (Answer is D)

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17 Dec 2025, 09:34

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Positive integers greater than 10 that have no factor p such that 1 < p < the generated number; means p is a prime number greater than 10

p = {11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,...}
18 = 2*3^2

p = 18k + r : where r can not be even or a multiple of 3

Remainders = {11,13,17,1,5,7} : 6 possible remainders since remainder is odd and 3,9,15 can not be remainder otherwise 3 will be a factor of p.

IMO D

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17 Dec 2025, 09:40

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Bunuel

Possible reminders = 1, 2, 3, 4, 5, 6, 7, 8, 9 , 10, 11, 12, 13, 14, 15, 16 , 17

The remainder can't be even the number generated is odd. Hence we can eliminate all even numbers.

Between 1, 3, 5, 7, 9, 11, 13, 17
If the remainder is 3 or 9, the number will be of form 18x + 3 or 18x + 9. In this case, the number would not be a prime number. Hence, we can eliminate 3 and 9.

Possible remainers = 1, 5, 7, 11, 13, 17

Option D

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17 Dec 2025, 09:41

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Possible odd remainders mod 18---1,3,5,7,9,11,13,15,17
removing multiples of 3---(3,9,15)
total number -6

Bunuel

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17 Dec 2025, 09:42

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A positive integer having no factor p between p and the integer itself means such integer is a prime number.

So, generated numbers are prime numbers greater than 10.

Generated number/18 => Remainder noted

Starting with 11

11/18 => Rem= 11
13/18 => Rem=13
17/18 => Rem=17
19/18 => Rem=1
23/18 => Rem=5
29/18 => Rem=11

Going on the same remainders will repeat.

So distinct remainders= 11,13,17,1,5

So, Number of distinct remainders=5

Option C

Bunuel