Weird Ones

Logic Level 3

$\LARGE 31 \\ \LARGE 331 \\ \LARGE 3331 \\ \LARGE 33331 \\ \LARGE 333331 \\ \LARGE 3333331 \\ \LARGE 33333331 \\ \LARGE 333333331 \\$

Which one of these numbers is not like the others?

3333331 331 3331 333333331 31 33331 33333331 333331

5 solutions

Pi Han Goh
May 6, 2015

All of these numbers are prime except the last one. $333333331 = 17 \times 19607843$ .

Challenge student note: Nice observation. Keep posting such beautiful problems.

Nihar Mahajan - 6 years, 1 month ago

hahaha @Nihar Mahajan (challenge student note).

Aditya Kumar - 6 years, 1 month ago

The next prime of this form has $17$ three's followed by a $1.$ :P

How about if we put other digits in place of $3$ ?

Brian Charlesworth - 6 years, 1 month ago

Prove that there's infinite number of primes of this form!

Pi Han Goh - 6 years, 1 month ago

Perhaps a good starting point is to observe that these numbers are of the form $\dfrac{10^n - 7}{3}$ for integers $n \gt 1.$

Brian Charlesworth - 6 years, 1 month ago

I didnt test all of them for primes and guessed that 31 is the only prime among them and marked 31 as the answer. And when i saw the solution i blamed my laziness.

Aditya Chauhan - 5 years, 9 months ago

I did the same thing

N T - 5 years, 6 months ago

I did just the opposite. I guessed the last number because I looked at the number of 3's and the factors of the number of 3's. All factors were prime except the last one. Basically I factored out the total number of 3's in each number and all the factors were prime numbers, except the last one that had 4 (as 8=4x2).

Ashton Hettich - 5 years, 6 months ago

Excuse me for saying this but how the hell on the earth are we supposed to factorize all those numbers?! and by the way this question is nothing even close to logic the way you solved it.

Arian Tashakkor - 6 years, 1 month ago

It's very difficult to factor them. But you can easily show that the first 3 numbers are prime. Then you could make the assumption that all are primes, testing out the primality of the rest of the numbers for potential small prime factors show that 17 is a factor of the last number. I posted this question in logic because it's an old gem in recreational mathematics and there isn't one right answers to sequences like these. That's why I posted this in Logic rather than in other topics whereby you need rigorous proof for everything.

Pi Han Goh - 6 years, 1 month ago

@Pi Han Goh Thank you sir.And BTW since that assumption can be a little hard to make (primality) don't you think the problem is slightly underrated?

Arian Tashakkor - 6 years, 1 month ago

@Arian Tashakkor – Hahaha, possibly. But I don't think sequences-related problems deserve a high rating in the first place.

Pi Han Goh - 6 years, 1 month ago

I answered 31... because 31 is not like the other numbers which are preceded by 3. :-)

Jay-Andrewson Dayuday - 5 years, 6 months ago

hi,just think of this solution......all the numbers ......when their digits are added (only 3 )every number gives some value at the units place of the result,but 7 3's number gives 21 giving 1 in units place .that matches with the left out 1... of every number.

Eswar Gvs - 6 years, 1 month ago

Nguyễn Nam
Aug 16, 2015

Another "weird one" solution of mine. I'm not really good at math so this is how I did:

3+1=4
3+3+1=7
3+3+3+1=10
3+3+3+3+1=13
3+3+3+3+3+1=16
3+3+3+3+3+3+1=19
3+3+3+3+3+3+3+1=22
3+3+3+3+3+3+3+3+1=25

Of course there would be odd numbers and even numbers alternatively. I separated all the numbers into 2 groups: even and odd. Nothing's strange in the group of even number, all of them are not prime. However, in the group of odd number only the last one is NOT prime. That's why I chose 333333331.

Actually I myself feel that this is not a good answer tho. Anyway (luckily) it helps me picking the right one.