Three different digits are chosen from among 1 - 9. When arranged in each of the six possible orders, each of the resulting 3-digit numbers is the product of exactly two primes

What are the three digits?

I found a simple process to solve this problem using nothing but
mental math and a list of primes under 1000.  It amounts to eliminating any
combinations that would have a either a permutation that is prime, or a permutation that contains more than 2 prime factors.  For the only way for a number to be equal to the product of 2 primes is if those 2 are its only prime factors.

My method has 3 parts.  First I use various divisibility short-cuts to eliminate combinations that might have more than 2 prime factors.  Then I eliminate those with permutations that are prime.  Then last I eliminate one of the remaining 2 by simple trial and error, thus leaving us with the only solution.

First start by listing all the ways of picking 3 digits from 1 to 9:

123,124,125,126,127,128,129,134,135,136,137,138,139,145,146,147,
148,149,156,157,158,159,167,168,169,178,179,189,234,235,236,237,
238,239,245,246,247,248,249,256,257,258,259,267,268,269,278,279,
289,345,346,347,348,349,356,357,358,359,367,368,369,378,379,389,
456,457,458,459,467,468,469,478,479,489,567,568,569,578,579,589,
678,679,689,789

now we can eliminate all combinations whose digits sum to a multiple of 3 and have an even digit.  This leaves us with:

124,125,127,128,134,135,136,137,139,145,146,148,149,157,158,159,
167,168,169,178,179,235,236,238,239,245,247,248,256,257,259,268,
269,278,289,346,347,349,356,357,358,359,367,368,379,389,457,458,
467,469,478,479,568,569,578,579,589,679,689

Next we an eliminate all combinations that sum up to 9 or 18 because then it would be a multiple of 9 and thus have more than 2 prime factors.

124,125,127,128,134,136,137,139,145,146,148,149,157,158,159,167,
168,169,178,179,235,236,238,239,245,247,248,256,257,259,268,269,
278,289,346,347,349,356,357,358,359,367,368,379,389,457,458,467,
469,478,479,568,569,578,579,589,679,689

Next we can eliminate all combinations that sum up to a multiple of 3 and contain the digit 5.  This leaves us with:

124,125,127,128,134,136,137,139,145,146,148,149,157,158,167,168,
169,178,179,235,236,238,239,245,247,248,256,257,259,268,269,278,
289,346,347,349,356,358,359,367,368,379,389,457,458,467,469,478,
479,568,569,578,589,679,689

Next we can eliminate all combinations that contain 2 digits that can form a multiple of 4.  Thus one of the six would be a multiple of 4.  Thus we are left with:

134,137,139,145,149,157,158,178,179,235,347,349,358,359,379,389,
457,479,578,589

Next we can eliminate any comination where one of the 6 permutations would be prime.  This leaves us with:

158,178

Now a little trial and error on 158 can show that 518=2*7*37.  Thus we are left with 178.  Now it can be verified that 178 works with
178=2*89
187=11*17
718=2*359
781=11*71
817=19*43
871=13*67

 

 

 

 

 

Posted by Daniel on 2009-08-15 23:10:39