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  • Writer's pictureNiedhie

Conjecture 21

Updated: Dec 23, 2022

Sometime in the past I had read or heard that it takes 21 days for a person to get rid of any of his habits. Just now I was thinking about it and was trying to figure out why was 21 chosen. What was wrong with any other number, was 21 just an arbitrary choice or some scientifically calculated estimate. I still do not have an answer to this question, but while thinking about this I have arrived at a nice discovery though.

Conjecture: 21, 42 and 84 and their reverses of the digits i.e 12, 24 and 48 are the only two digit numbers (so that both digits are non-zero and exclude the multiples of 9) which are divisible by what their digits sum up to.

Proof: Let us try to prove this conjecture. I say it as a conjecture as I am still to get a complete and adequate proof to it. But logic and intuition makes me say that it is provable.

The only reason to exclude 1 to 9 is that they trivially hold as the sum of digits in this case is just the number which is surely divisible by itself. Also the numbers with multiples of 10 add up to the non-zero digit which would surely divide the number. Also the multiples of 9 have this property that they sum up to 9 and hence would surely be divisible by 9. So let’s concentrate on this more interesting case of 2 digit numbers i.e., from 11 to 99.

Certainly the first thought would be to exclude all the prime numbers in this range as they can never be divided by their sum.

Next we can exclude even numbers with an odd in the first digit, as they would certainly add up to an odd number which can never divide the number which is even. So this excludes all the even numbers in those columns with first digit as odd e.g columns of 10s, 30s etc.

Also exclude numbers with both digits odd as they would sum to even which can not divide the odd number.

Next the numbers with repeated digits on being divided by their sum would always give 11/2 which means they are not divisible. So exclude them.

The numbers ending with 5 can only be divided by 5 or 0 and any non zero digit added to 5 can never give 5, so exclude all numbers ending with 5.

The rest can be excluded by hit and trial.

This is surely not the greatest way to solve this, the best way would be to make an algorithm and program this rather than solving it so inefficiently as I suggested. Now since I have already typed it by putting in so much of efforts I would surely post it and welcome any suggestion on this problem also, but you are equally free to simply ignore this post.

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