# double the number

Arjun Shastry

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Posts: 1898

1

posted 5 years ago

This one is quick and interesting.

Natural number ends with 2.If we remove this 2 and place it at the beginning, the number gets doubled.Find the smallest possible original number.

(e.g. number is say 142. New number will be 214. 214 is not twice 142 so 142 is not such number).

Natural number ends with 2.If we remove this 2 and place it at the beginning, the number gets doubled.Find the smallest possible original number.

(e.g. number is say 142. New number will be 214. 214 is not twice 142 so 142 is not such number).

MH

palanisamy subramani

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Mike Simmons

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Arjun Shastry

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Posts: 1898

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anirudh jagithyala

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Arjun Shastry

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Posts: 1898

1

posted 5 years ago

Maybe you could interpret the title as 'diversions AWAY from programming', thus no programming required.

Paul Clapham wrote:No need to wait... the title of this forum isProgrammingDiversions. While the question is certainly diverting, I haven't seen any programming yet.

Maybe you could interpret the title as 'diversions AWAY from programming', thus no programming required.

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Mike Simmons

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posted 5 years ago

This gives all solutions up to 100 digits:

The result is

The pattern is clear.

Obviously, what I was solving for was all the digits other than the 2, which moves.

The result is

The pattern is clear.

Obviously, what I was solving for was all the digits other than the 2, which moves.

Eugene Rabii

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Posts: 30

posted 5 years ago

-4? -2 surely?

My thinking runs this way:

2 * (10^n) + (x-2)/10 = 2x

2 * (10^(n+1)) + x - 2 = 20x

2 * (10^(n+1)) - 2 = 19x

x = (2 * (10^(n+1)) - 2 ) / 19

So, we just look for a value of n that makes x an integer ending with 2. The n vs (n+1) doesn't matter, but the -4 vs -2 will produce different answers.

OK, code time:

Results:

2, 10, 8

3, 105, 3

4, 1052, 10

5, 10526, 4

6, 105263, 1

7, 1052631, 9

8, 10526315, 13

9, 105263157, 15

10, 1052631578, 16

11, 10526315789, 7

12, 105263157894, 12

13, 1052631578947, 5

14, 10526315789473, 11

15, 105263157894736, 14

16, 1052631578947368, 6

17, 10526315789473684, 2

18, 105263157894736842, 0

So the answer is 105263157894736842. The check for the last digit being 2 turned out to be unnecessary, but I

ETA: In reading through Mike's solution, I see the -4 makes sense if you declare x to be the number without the final 2. Sorry about that!

I wonder if we can prove we don't have to check the final digit, as both Mike and I discovered?

Eugene Rabii wrote:I have to say, this is impressive!

I have reached the same code:

x = ( 2*(10^n) - 4 ) / 19

And it was a pleasure getting here!

Cheers, Eugene

-4? -2 surely?

My thinking runs this way:

2 * (10^n) + (x-2)/10 = 2x

2 * (10^(n+1)) + x - 2 = 20x

2 * (10^(n+1)) - 2 = 19x

x = (2 * (10^(n+1)) - 2 ) / 19

So, we just look for a value of n that makes x an integer ending with 2. The n vs (n+1) doesn't matter, but the -4 vs -2 will produce different answers.

OK, code time:

Results:

2, 10, 8

3, 105, 3

4, 1052, 10

5, 10526, 4

6, 105263, 1

7, 1052631, 9

8, 10526315, 13

9, 105263157, 15

10, 1052631578, 16

11, 10526315789, 7

12, 105263157894, 12

13, 1052631578947, 5

14, 10526315789473, 11

15, 105263157894736, 14

16, 1052631578947368, 6

17, 10526315789473684, 2

18, 105263157894736842, 0

So the answer is 105263157894736842. The check for the last digit being 2 turned out to be unnecessary, but I

*think*that's a coincidence.

ETA: In reading through Mike's solution, I see the -4 makes sense if you declare x to be the number without the final 2. Sorry about that!

I wonder if we can prove we don't have to check the final digit, as both Mike and I discovered?

Mike Simmons

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Posts: 3090

14

posted 5 years ago

I tried some thing very simple and may sound very stupid....

BigInteger i,a,b;

for(i=0;i<100000000000000;i++)

{

String x,y;

String num = Integer.toString(i);

x = num+"2";

y = "2"+num;

int a = Integer.parseInt(x);

int b = Integer.parseInt(y);

if(a==(b/2))

System.out.println(a+" "+b);

}

did not get any no...though....

BigInteger i,a,b;

for(i=0;i<100000000000000;i++)

{

String x,y;

String num = Integer.toString(i);

x = num+"2";

y = "2"+num;

int a = Integer.parseInt(x);

int b = Integer.parseInt(y);

if(a==(b/2))

System.out.println(a+" "+b);

}

did not get any no...though....

SCJP