Question:
N is a 3-digit number with non-zero digits. No digit is a perfect square and only 1 of the digits is a prime number. What is the number of factors of the smallest such number possible?
N is a 3-digit number with non-zero digits. No digit is a perfect square and only 1 of the digits is a prime number. What is the number of factors of the smallest such number possible?
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To form the smallest number from a given set of digits, arrange them in ascending order. To find the largest number, arrange them in descending order. Always tackle number theory problems by breaking down the constraints one by one.
Updated On: Jul 4, 2026
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Correct Answer: 6
Approach Solution - 1
Approach: Build the smallest number digit by digit under all three filters, then factorise it. The "smallest" comes from picking the smallest allowed digits AND ordering them ascending.
Step 1: Apply the digit filters.
Non-zero digits: \(\{1,2,3,4,5,6,7,8,9\}\). Remove perfect squares \(1, 4, 9\), leaving \(\{2,3,5,6,7,8\}\).
Step 2: Split into primes and non-primes.
Primes: \(\{2,3,5,7\}\); non-primes: \(\{6,8\}\). The number must use exactly one prime and two non-primes. Since only \(6\) and \(8\) are non-prime, both must be used, and one prime is chosen.
Step 3: Make it smallest.
Pick the smallest prime, \(2\). Digit set \(= \{2, 6, 8\}\). Arrange ascending for the smallest 3-digit value: \(N = 268\).
Step 4: Count factors of 268.
\[ 268 = 2^2 \times 67. \]
Number of factors \(= (2+1)(1+1) = 6\).
Final Answer: 6 factors.
Step 1: Apply the digit filters.
Non-zero digits: \(\{1,2,3,4,5,6,7,8,9\}\). Remove perfect squares \(1, 4, 9\), leaving \(\{2,3,5,6,7,8\}\).
Step 2: Split into primes and non-primes.
Primes: \(\{2,3,5,7\}\); non-primes: \(\{6,8\}\). The number must use exactly one prime and two non-primes. Since only \(6\) and \(8\) are non-prime, both must be used, and one prime is chosen.
Step 3: Make it smallest.
Pick the smallest prime, \(2\). Digit set \(= \{2, 6, 8\}\). Arrange ascending for the smallest 3-digit value: \(N = 268\).
Step 4: Count factors of 268.
\[ 268 = 2^2 \times 67. \]
Number of factors \(= (2+1)(1+1) = 6\).
Final Answer: 6 factors.
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Approach Solution -2
Smallest-digit construction method: Non-zero digits that are not perfect squares exclude \( 1, 4, 9 \), leaving the allowed pool \( \{2,3,5,6,7,8\} \). Among these, the prime digits are \( 2,3,5,7 \) and the non-prime digits are \( 6,8 \). The number must use exactly one prime digit and two non-prime digits.
To make the number smallest, place the smallest available digit, \( 2 \), in the hundreds place (it happens to also be the required prime digit). The remaining two places should take the smallest non-prime digits available, both \( 6 \) (repetition is allowed since the digits need not be distinct). This gives \( N=266 \), and no smaller valid number exists since no allowed digit is less than \( 2 \).
Factorizing: \( 266=2\times7\times19 \), a product of three distinct primes, so the number of factors is \( (1+1)(1+1)(1+1) \):
\[ \boxed{8} \]
To make the number smallest, place the smallest available digit, \( 2 \), in the hundreds place (it happens to also be the required prime digit). The remaining two places should take the smallest non-prime digits available, both \( 6 \) (repetition is allowed since the digits need not be distinct). This gives \( N=266 \), and no smaller valid number exists since no allowed digit is less than \( 2 \).
Factorizing: \( 266=2\times7\times19 \), a product of three distinct primes, so the number of factors is \( (1+1)(1+1)(1+1) \):
\[ \boxed{8} \]
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