### Brain Teasers

# 50% Larger

Find the smallest number such that if its rightmost digit is placed at its left end, the new number so formed is precisely 50% larger than the original number.

### Answer

The answer is 285714.If its rightmost digit is placed at its left end, the new number is 428571 which is 50% larger than the original number 285714.

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## Comments

The instructions were unclear to me as to whether I was to "move" the last digit to the front or rather "add" the last digit to the front.

Just "adding" the last digit to the front has

no solutions, because the digit that would

have to be added is 1 (otherwise you increase

by more than 50%), and that means that the original

integer is odd, so you can't increase by exactly

50% to get another integer. A possible exception

is 0, but "00" is not a standard integer representation

(unless we are talking about roulette or basketball).

So the teaser must mean moving the last digit to the

front, which is the only way it makes sense.

no solutions, because the digit that would

have to be added is 1 (otherwise you increase

by more than 50%), and that means that the original

integer is odd, so you can't increase by exactly

50% to get another integer. A possible exception

is 0, but "00" is not a standard integer representation

(unless we are talking about roulette or basketball).

So the teaser must mean moving the last digit to the

front, which is the only way it makes sense.

er..isn't moving the rightmost digit to the left end, exactly the same as "moving the last digit to the front"???

You said the rightmost digit was "placed"

at the left end, which leaves it somewhat

ambiguous whether the original digit

or a copy was so placed. However, since

the copy interpretation doesn't lead to any nontrivial

solutions, it must have been the original

that was moved.

at the left end, which leaves it somewhat

ambiguous whether the original digit

or a copy was so placed. However, since

the copy interpretation doesn't lead to any nontrivial

solutions, it must have been the original

that was moved.

It does say "rightmost digit IS placed at its left end" not "its rightmost digit is duplicated or copied at its left end, ".

I personally can not see where there could be any confusion. It does quite clearly state the THE right most number is moved (placed) from the right to the left end, NOT duplicated or copied, but quite clearly Moved(placed).

Ambigous doesn't even come into it?

I personally can not see where there could be any confusion. It does quite clearly state the THE right most number is moved (placed) from the right to the left end, NOT duplicated or copied, but quite clearly Moved(placed).

Ambigous doesn't even come into it?

No answer is correct. You ask for the smallest number so 0.285714 would work but then 0.0285714 also works and is smaller, but 0.00285714 also works... (you see my point)

I think you should add "whole integer" to your instructions somewhere. But otherwise a great teaser which is quite possible impossibly hard for non-rainman types, like me, to solve. 8^)

I think you should add "whole integer" to your instructions somewhere. But otherwise a great teaser which is quite possible impossibly hard for non-rainman types, like me, to solve. 8^)

I don't get your point at all, EJ. 40.28571 is NOT 50% larger than 0.28571; even if you replace the 0 with a 4, 4.28571 is NOT 50% larger than 0.28571 either.

You can't just let the decimal point float anywhere you want it too!

This will only work with whole numbers.

You can't just let the decimal point float anywhere you want it too!

This will only work with whole numbers.

I stand corrected. I reread the way this teaser is worded and you are right, it is fine. I was thinking about moving the last digit to the first significant digit (which is not what this teaser says) so 0.428571 is 50% larger than 0.285714 and 0.0428571 is 50% larger than 0.0285714. But as you pointed out the way this is worded would put the last digit all the way to the right and make it 40.028571 in the last case. My bad.

dosent 50% larger mean one half of the number???????????? or maby I'm just dumb.

Er..half of 285714 is 142857, and the difference between 428571 and 285714 is 142857

wow.

It was a good teaser; the math was a bit involved, but the teaser was well written and good just the same!

First, let me say: COOL teaser!

However, neither the given answer nor any of these comments have explained how to find this number.

After thinking about the problem for a couple of minutes (kinda embarrassed it took that long to get it), I went directly to the answer, no trial and error. Further, I can also tell you that the next lowest number with this property is 571428 and that there are no other numbers less than 10^50 with this property.

What I realized is that I needed to find a repeating fraction such that the number of repeating digits was one less than the number. This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e. every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).

Now I wanted to find any occurances where the nth digit followed the 1.5nth digit.

The repeating digits in 1/7 and their corresponding 'n' where the digits are the first digit after the decimal point are:

142857 - repeating digits

132645 - n

n=2 follows n=3, so the lowest number is 285714. Also, n=4 follows n=6, so the next number is 571428.

Out of curiosity, I checked all the prime numbers up to 50. The ones with p-1 repeating digits are:

17, 19, 23, 29, 31, 47

None of these has two adjacent digits with the desired property, so there is no other solution with less than 50 digits.

I'd be curious if there was a way to either prove that no other solutions exist, or determine which prime numbers would have the required relationship. I think other solutions do exist, but I'll write a program to check them any furthur.

However, neither the given answer nor any of these comments have explained how to find this number.

After thinking about the problem for a couple of minutes (kinda embarrassed it took that long to get it), I went directly to the answer, no trial and error. Further, I can also tell you that the next lowest number with this property is 571428 and that there are no other numbers less than 10^50 with this property.

What I realized is that I needed to find a repeating fraction such that the number of repeating digits was one less than the number. This would be a prime number, and seven is the lowest number with this property. With this property, every digit in the repeating fraction appears in each place exactly once (i.e. every repeated digit appears as the first digit after the decimal exactly once for n/p where p > n > 0).

Now I wanted to find any occurances where the nth digit followed the 1.5nth digit.

The repeating digits in 1/7 and their corresponding 'n' where the digits are the first digit after the decimal point are:

142857 - repeating digits

132645 - n

n=2 follows n=3, so the lowest number is 285714. Also, n=4 follows n=6, so the next number is 571428.

Out of curiosity, I checked all the prime numbers up to 50. The ones with p-1 repeating digits are:

17, 19, 23, 29, 31, 47

None of these has two adjacent digits with the desired property, so there is no other solution with less than 50 digits.

I'd be curious if there was a way to either prove that no other solutions exist, or determine which prime numbers would have the required relationship. I think other solutions do exist, but I'll write a program to check them any furthur.

I agree with earlier comments that it should be "moved" instead of "placed". There are some excellent explanations here of why the first interpretation is impossible and a nice explanation by the Guru. What lead to the realisation that it was a fraction with repeating digits? I see it but I don't get why. I think I'll PM the Guru.

Thanks for the teaser.

After formulating the problem as a bunch of linear equations, I used scip (a linear, mixed integer and nonlinear programming solver) to find the solution. Not knowing beforehand whether the integer solution had more digits than scip (actually the host OS math library) could handle, I had to teach it long division and long addition. A bit tedious, but it surely worked! I also found out that repeated patterns of the solution (e.g. 285714285714) also match the criteria (but obviously are not the "smallest number").

The model file that I used is available at pastebin for anyone who might be interested to play with it. https://pastebin.com/V3z8TzLu

After formulating the problem as a bunch of linear equations, I used scip (a linear, mixed integer and nonlinear programming solver) to find the solution. Not knowing beforehand whether the integer solution had more digits than scip (actually the host OS math library) could handle, I had to teach it long division and long addition. A bit tedious, but it surely worked! I also found out that repeated patterns of the solution (e.g. 285714285714) also match the criteria (but obviously are not the "smallest number").

The model file that I used is available at pastebin for anyone who might be interested to play with it. https://pastebin.com/V3z8TzLu

I just found yet another way to tackle the problem. It just so happens that R has a library (rcdd) that can solve linear programming problems (such as this one) with very high precision (using rational arithmetic).

This allows for a very concise program formulation and blazingly fast solutions.

If you are interested, the tiny (15-line) R program and a shell script wrapper, which can find solutions for arbitrary number sizes, are available at https://pastebin.com/FMAuTWV7

This allows for a very concise program formulation and blazingly fast solutions.

If you are interested, the tiny (15-line) R program and a shell script wrapper, which can find solutions for arbitrary number sizes, are available at https://pastebin.com/FMAuTWV7

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