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

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Rahilgaur

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17 Dec 2025, 09:44

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Bunuel

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

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Since, generated number cannot have a factor that is less than 10 it must be a prime number.

Upon dividing any number by 18, we can have remainder from 1 to 17---> 17 remainders

Factor of 18 are 2, 3, 1, 18

1. Even remainders - eliminate all even remainders since 18x+even cannot be a prime number. (2,4,6,8,10,12,14,16) - 9 remainders remaining

2. Remainder multiple of 3 - eliminate since 18x+3y will always be divisible by 3 and cannot be prime - (3,6,9,12,15 ) - 6 remainders remaining (1, 5,7,11,13,17) .. Answer D. 6

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The number has no factor p meaning it is a prime number and number >10
The number is divided by 18 and remainder is recorded
Prime factors of 18 are 2 and 3
2 and 3 can't divide any prime number other than 2 and 3
The number is a co prime of 18
Therefore, the only possible remainders are 1,5,7,11,13,17. (Numbers which are not divisible by 2 or 3)

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17 Dec 2025, 09:53

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IMO answer is C - 5
basically they are saying generated numbers are prime numbers as no factores between 1 and no. generated.
So prime factors greater than 10 divided by 18 remainder are
11 : 11
13 : 13
17 : 17
19 : 1
23 : 5
29 : 11 repeates
31: 13
Hence remainders are 11,13,17,1,5 so 5

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17 Dec 2025, 10:00

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Essentially, the program generates prime numbers greater than 10.
We will take a few to find a pattern here
11, 13, 17, 19, 23, 29, 43, 47, 53, 59

All these give remainders as either 11, 13, 17, 1, 5, or 7 when divided by 18.
So a total of 6.
Option D.

Bunuel

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ghimires28

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17 Dec 2025, 10:09

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lets note down prime number grater and 10 and see if we can find a pattern
11,13,17,1923,29,31...
number reaminder
11 11
13 13
17 17
19 1
23 5
29 11
31 13
41 5
5 numbers

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17 Dec 2025, 10:11

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From the question stem, it is clearly understand that it is a prime number.

Prime number is in the form of 6k+/-1 excluding 2 and 3.

Given that positive numbers are greater than 10 that means we can eliminate 2&3, which doesn't fit the form 6k+/-1.

6k+/-1 can be written as 6(1)+/-1, 6(2)+/-1, 6(3)+/-1, 6(4)+/-1 and so on.... => 5,7,11,13,17,19,23,25,29,31,35,37......

One can see that after 5,7,11,13,17,19 all number are can be written with adding 18,36,54.... (18 multiple) to the set of numbers mentioned in the beginning.

18,36,54... are multiples of 18, so the reminder when 6k+/-1 series is divided by 18 would be 5,7,11,13,17,1.

So, the number of unique remainders would be 6.

Bunuel

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17 Dec 2025, 10:12

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Bunuel

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no factor p is the biggest clue that the numbers are prime numbers.
Prime numbers greater than 10 will be (11,13,17,19,23...)

When we divide these prime numbers with 18 - n=18k+r where k is the quotient and r is the remainder. maximum Possible remainders will be (0,1,2,3..17)
N is a number greater than 10 and prime so we will have to eliminate remainders which have common factors with 18 and r. hence following remainders are possible
(1,5,7,11,13,17).
hence 6

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17 Dec 2025, 10:19

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The answer is 5 (Distinct Remainders are : 11,13,17,1 and 5) - as positive integer greater than 10 - that have no factor p such that 1<p<generated numer means the number is prime therefore divide primes such as 11, 13, 17,19 etc by 18

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17 Dec 2025, 10:23

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The program generates positive integers that are greater than 10 and have no other factors other than 1 and itself. Those numbers could be 11, 13,17,19,23,29,31,37,41........
Any number divided by 18 can only leave a remainder between 0-17.
18= 2*3*3
A prime number greater than 10 cannot be even, so we can eliminate even remainders (0,2,4,6,8,10,12,14,16) and also it cannot be divisible by 3, so we can eliminate multiples of 3 (0,3,6,9,12,15)
Now remaining we have only 1,5,7,11,13,17
We have 6 distinct possible remainders.

D

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17 Dec 2025, 10:36

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The question clearly refers to Prime numbers > 10. Thus, the possibilities are 11, 13, 15, 17, 19, 23, 29, 31... The remainder, after dividing by 18, are 11, 13, 17, 1, 4, 11, ...

The number of distinct remainders are 5.

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The only way in which a number has factors only as 1 and the number itself is when the number is prime. A prime number can be of the form 6k+1 or 6k-1 and we need to find the remainder when these numbers are divided by 18.

Trying out various prime numbers greater than 10, we can see the remainders are always 1,5,7,11,13, and 17.

Therefore, Option D

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The question is asking the number of distinct remainders when prime numbers greater than 10 are divided by 18.

we know that all primes are of the form 6q+-1. Plug in values such that q is greater than 1 (as primes greater than 10 is asked).
Also note that any number divided by 18 can have remainders (0-17).

Plug in values till you get remainder 17. Hence remainders can be 1,5,7,11,13,17.

Total of 6 values. Hence answer is D

Bunuel

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In short the question asks for prime numbers above 10 and what unique reminders we can get by dividing all the prime above 11 with 18.
Since they are few in number as shown by the answers we can just list them down and test them
11/18 = rem 11
13/18= rem 13
17/18= rem 17
19/18= rem 1
23/18= rem 5
43/18= rem 7
Total 6
Ans D

Bunuel

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17 Dec 2025, 11:10

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11/18 R=11
13/18 R=13
17/19 R=17
19/18 R=1
23/18 R=5
this is will follow for all the prime numbers hence Ans IMO C

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Bunuel

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Each prime divided by 18 can have a remainder: {1,2,3,4,5....17}
Remainder not a multiple of 2 & 3, so remove them: {0,2,3,4,6,8,9,10,12,14,15,16}
Remaining: {1,5,7,11,13,17}
Total = 6

Hence, OPTION D.

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p is prime above 10
remainder 1-17
now, p= 18k+r
for p to be odd, r has to be odd
if r is 3, 9,15..p cannot be prime.
so possible remainders 1,5,7,11,13,17

total 6
Ans D

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17 Dec 2025, 11:40

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Let the generated positive integer be x

Given: x>10 and there is no factor p such that 1<p<x (which implies x is a prime number)

We have to identify remainders when x/18

Computing remainders for first few cases:

11, 13, 17 - which when when divided by 18 leaves the same number as remainder.
19 mod 18 = 1
23 mod 18 = 5
29 mod 18 = 11
31 mod 18 = 13
37 mod 18 = 1
41 mod 18 = 5
43 mod 18 = 7
47 mod 18 = 11
53 mod 18 = 17
59 mod 18 = 5

This shows that there are 6 possible values coming in different order.

Hence, the answer is (D)

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The numbers described are prime numbers greater than 10.( p have no factor)

When a prime number is divided by 18 its remainder cannot share a common factor, otherwise the prime would be divisible by 2 or 3. So valid remainders are those 1,5,7,11,13,17 total 6

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17 Dec 2025, 11:59

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Bunuel

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The inequality 1 < p < # generated indicates that the generated numbers are always prime, because they have no factors that are less than themselves other than 1. Let's check the prime numbers greater than 10 and see if we can find a pattern.

$\frac{11}{18}= R11$

$\frac{13}{18}= R13$

$\frac{17}{18}= R17$

$\frac{19}{18}= R1$

$\frac{23}{18}= R5$

$\frac{29}{18}= R11$

$\frac{31}{18}= R13$

Pattern is repeating now, we got 5 different values for remainder. Thus answer is C, 5.

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17 Dec 2025, 12:25

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Bunuel

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We need to generate a random positive number greater than 10.

It should not contain any factor greater than p. Such that 1 < P < random number.

If there exists no factor for the random number generated apart from the number and 1.

It’s obvious, the random number generated is a prime number.

So, prime numbers greater than 10 are : 11,13,17,19,23,29,31,37,43,47 etc.

So if, we divide each number by 18, the remainder we get are as follows:

11 ➗ 18 , we get remainder as 11.

13 ➗ 18 , we get the remainder as 13.

17 ➗ 18, we get the remainder as 17.

19 ➗ 18, we get the remainder as 1.

23 ➗ 18, we get the remainder 5.

29 ➗ 18, we get the remainder 11.

31 ➗ 18, we get the remainder 13.

37 ➗ 18, we get the remainder 1.

43 ➗ 18, we get the remainder 7.

The distinct remainders are : 1,5,7,11,13 and 17.

6 cases.

Alternate approach:

Prime numbers are of the form 6k+1 or 6k-1. What’s the remainder when the number is divided by 18.

If K =3, we get 19 and 17.

The remainders are 1 and 17.

If K= 4, we get 25 and 23.

25 is not a prime. Hence, the remainder is 5.

If k=5, we get 31 and 29.

The remainders we get are : 13 and 11.

If k=6, we get repetitive of remainders.

If k=7, the numbers we get are 43 and 41.

The remainders are 7 and 5.

The distinct remainders are : 1,17,5,13,11,7.

Totally 6.

The question arises where do we stop, as we have an option as 9. Do we need to iterate even more ?

Number generated is of the form 18K+N

And the number N should be coprime with 18.

Why? 18K +2 = 2(9k+1) = multiple of 2. So cannot be a prime number.

Coprime with 18 are : 1,5,7,11,13,17.

So these are the 6 remainders.

Option D