# Question Of The Month: September

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Question Of The Month

“Find a set of three digits (not necessarily all different) such that whichever order one wrote them down. The resultant 3-digit number was a prime.”

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131

101

I think the only sets that meet that condition are 111-111-111 and 131-113-311. What’s the answer Ofili?

Ngo

113, 311, 131…it cant be 111, cuz those digits add up to three which means its divisible by three. It cant have any zero’s cuz then it’s divisible by 2 and 5, it can have even numbers because then it’s divisible by 2, it cant have 9 because it is divisible by 3. Nor can we use 5 cuz it’s always divisible by 5, so the only digits left to work with are 1, 3, and 7…..and working and twisting with those numbers, I got that 113, 311, and 131 are all prime.

311 & 111

1, 1, 3

and 3, 3, 7

Cezanne, you’re right, it’s only 113-131-311 and not 111, which equals 3×37. Thanks.

Ngo

Let me just say that there are 3 possible answers. And who ever comes up with an original/different set first could potentially win a book =)

I was wrong about the 9 in my original post. 179 197 and 971 and 733 337 373 and 113 131 and 311

1, 9, and 9 seems to work.

363

1, 3, and 7 i think… any order you put them in you will get a prime number. Other wise I did not understand the question and for that I will say: “D’oh! I tried!” 😉

–Edgar

@Edgar: 31 X 23 = 713 (not a prime)

@Cezanne: 7 X 131 = 917

199, 919, 991 …final answer, if this is wrong…idk…this goes with the [1-3-1] combo and the [3-7-3] but Jr Saul got it first :'(

Considering the numbers 1,1, and 3 offered only three unique solutions, another set of digits, 3, 7, and 9, can produce four unique primes: 379, 937, 739, and 397.