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12 Jan 2026, 06:56
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A computer program randomly generates positive integers greater than 10 that have no factor \(p\) such that \(1 < p < \text{the generated number}\). 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
A. 3
B. 4
C. 5
D. 6
E. 9
12 Jan 2026, 06:56
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Official Solution:
A computer program randomly generates positive integers greater than 10 that have no factor \(p\) such that \(1 < p < \text{the generated number}\). 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
A generated number not having a factor \(p\) such that \(1 < p < \text{the generated number}\) 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share any factor greater than 1, then we could factor that out from \(18q + remainder\), and the expression would no longer represent a two or more digits prime number number.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
A computer program randomly generates positive integers greater than 10 that have no factor \(p\) such that \(1 < p < \text{the generated number}\). 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
A generated number not having a factor \(p\) such that \(1 < p < \text{the generated number}\) 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
