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17 Dec 2025, 23:27
Kudos
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Question says it should be greater than 10 and have no factor common other than it self and 1 , so it clearly indicated the property of Prime No.
so we can trial and error for all prime no greater than 10 and divide it by 18 to see what reminders are we getting repeatedly
11/18=R(11)
13/18=R 13
17/18= R 17
19/18= R 1
23/18= R 5
29/18=R 11(REPEATING FROM HERE)
SO we have 5 unique reminder
so ans is option C 5
so we can trial and error for all prime no greater than 10 and divide it by 18 to see what reminders are we getting repeatedly
11/18=R(11)
13/18=R 13
17/18= R 17
19/18= R 1
23/18= R 5
29/18=R 11(REPEATING FROM HERE)
SO we have 5 unique reminder
so ans is option C 5
Bunuel
msignatius
18 Dec 2025, 00:23
Kudos
Bookmarks
There's definitely a trend with the Christmas questions - quite tricky 
This one has two such "tricks" - It needs you to realize that if an integer greater than 10 has no factor between 1 and itself, it's a prime number. This one also needs you to keep on defining the remainder until 2 or 3 cycles, as there's a 6-remainder hidden a bit later.
I started from the prime numbers right after 10:
11 leaves a remainder of 11
13 leaves a remainder of 13
17 leaves a remainder of 17
19 leaves a remainder of 1
23 leaves a remainder of 5
29 leaves a remainder of 11
31 leaves a remainder of 13
37 leaves a remainder of 1
41 leaves a remainder of 5
43 leaves a remainder of 7
47 leaves a remainder of 11
53 leaves a remainder of 17
59 leaves a remainder of 5
61 leaves a remainder of 7
This is why I really went until 61 - you don't get the first instance of remainder-7 until 43, and if you stop before that, there are 2 cycles that complete without it. The reason is, there's that prime number around the multiple to 18 popping by in alternative cycles that shifts the equation just a bit.
Answer is 6.
This one has two such "tricks" - It needs you to realize that if an integer greater than 10 has no factor between 1 and itself, it's a prime number. This one also needs you to keep on defining the remainder until 2 or 3 cycles, as there's a 6-remainder hidden a bit later.
I started from the prime numbers right after 10:
11 leaves a remainder of 11
13 leaves a remainder of 13
17 leaves a remainder of 17
19 leaves a remainder of 1
23 leaves a remainder of 5
29 leaves a remainder of 11
31 leaves a remainder of 13
37 leaves a remainder of 1
41 leaves a remainder of 5
43 leaves a remainder of 7
47 leaves a remainder of 11
53 leaves a remainder of 17
59 leaves a remainder of 5
61 leaves a remainder of 7
This is why I really went until 61 - you don't get the first instance of remainder-7 until 43, and if you stop before that, there are 2 cycles that complete without it. The reason is, there's that prime number around the multiple to 18 popping by in alternative cycles that shifts the equation just a bit.
Answer is 6.
Bunuel
18 Dec 2025, 00:43
Kudos
Bookmarks
What numbers are generated ;
Numbers > 10
As mentioned no factor between 1 and itself
- These are prime numbers greater than 10 .
Examples : 11, 13 , 17 ,19 ......
What remainders are possible when divided by 18
0, 1,2....,17
But primes cannot be
even (divisible by 2 ) or divisible by 3 because 18 = 2 x 3^2
So survived remainders include 1,5,7,11,13,17
When you divide by 18 these numbers for example 19 , 23 , 43 , 11, 13 , 17 you get remainders as 1,5,7,11,13,17. So correct ans is 6
Numbers > 10
As mentioned no factor between 1 and itself
- These are prime numbers greater than 10 .
Examples : 11, 13 , 17 ,19 ......
What remainders are possible when divided by 18
0, 1,2....,17
But primes cannot be
even (divisible by 2 ) or divisible by 3 because 18 = 2 x 3^2
So survived remainders include 1,5,7,11,13,17
When you divide by 18 these numbers for example 19 , 23 , 43 , 11, 13 , 17 you get remainders as 1,5,7,11,13,17. So correct ans is 6
