## The Torn Number Puzzle

**BigThanks go out to Bilbao for submitting this one!**

I had the other day in my possession a label bearing the number 3 0 2 5 in large figures.

This got accidentally torn in half, so that 3 0 was on one piece and 2 5 on the other.

On looking at these pieces, I discovered this little peculiarity. If we add the 3 0 and the 2 5 together and square the sum we get as the result the complete original number on the label! Thus, 30 added to 25 is 55, and 55 multiplied by 55 is 3025.

Now, the puzzle is to find another number, composed of four figures, all different, which may be divided in the middle and produce the same result.

*Answers to this challenge can be entered into the section below; submissions will automatically be revealed when time is up!*

*More Puzzles From Bilbao:*

kkaa| Profile March 15th, 2011 - 12:46 am2025

kryz| Profile March 15th, 2011 - 7:18 am2025

Shawn| PUZZLE GRANDMASTER | Profile March 15th, 2011 - 8:05 amHow about 2025?

(20 + 25)^2 = 45^2 = 2025

Shawn| PUZZLE GRANDMASTER | Profile March 15th, 2011 - 8:11 am9801 works, too.

Shawn| PUZZLE GRANDMASTER | Profile March 15th, 2011 - 8:43 amI also find only 2 cases where this would work for an 8-digit number…

Vik| Profile March 15th, 2011 - 12:29 pm9801

Falwan| Profile March 16th, 2011 - 3:15 am9801====>98+01=99

99 × 99 = 9801

To tell the truth,I found it in the net.

suineg| PUZZLE MASTER | Profile March 16th, 2011 - 7:13 amAn easy answer to this is to say that the number is 2530, same digits but a total different number, I think this could be an answer but I will try to find a really different number with different digits, cool!!

munna| Profile March 16th, 2011 - 7:39 amAnswer = 9801

98 + 01 = 99

99 * 99 = 9801

suineg| PUZZLE MASTER | Profile March 16th, 2011 - 8:07 amOk I found the number, this is the real answer the other implies too much of wording tricks in the problem:

the number 9801 , splits 98 and 01 the sum is 99 and square of (99) is 9801, cool!!

The only thing I deduce from the problem rules is that the sum has to be a factor of 11, and that the sum of the digits to the right and the sum of the left digits of the parts has to be equal to the the sum of both parts divided by eleven, this an a little trial an error and I got it, cool man, nice puzzle Bilbao.

Adam| Profile March 16th, 2011 - 8:56 am9801

SubVet571| Profile March 16th, 2011 - 9:21 amI believe the answer is 2025 20+25=45 45*45=2025

Just my 2 cents worth

Steve

Bobo The Bear| PUZZLE MASTER | Profile March 16th, 2011 - 11:09 amIf the two 2 digit numbers are X and Y, we can write

X + Y = N, and

100X + Y = N^2, so

100X – X = N^2 – N

99X = N(N-1)

This equation was satisfied by X=30, N=55.

Another easy way to satisfy the equation is X=98, N=99

This yields X=98, Y=01, N=99, and

(98 + 01)^2 = 9801

(Also (20 + 25)^2 = 2025 is true, but violates the “four figures, all different” rule.)

John Lewis Highsmith| Profile March 16th, 2011 - 11:53 amThe value is 9801, which is the square of 99.

shaks| Profile March 16th, 2011 - 4:07 pmI think “9801″

ripped down the centre, you get 98 and 1 = 99, 99 squared is 9801

Good one!

virgolady85| Profile March 17th, 2011 - 12:59 am2025

20+25=45

45*45=2025

Mashplum| PUZZLE MASTER | Profile March 17th, 2011 - 9:49 pm9801

joe| Profile March 18th, 2011 - 9:30 am9801

98 + 1 = 99

99×99 = 9801

nice one Bilbao

cwallen| Profile March 18th, 2011 - 10:09 pm98 and 01

98*01 = 99

99^2 = 9801

RK| Founder | Profile March 19th, 2011 - 11:07 amHi there guys, glad to see you liked this one : )

The answer Bilbao had for this was 9801

Mashplum| PUZZLE MASTER | Profile March 20th, 2011 - 10:15 pm@Bilbao

I used Excel to figure this out by trial and error.

Is there a simple/clever/elegant way to solve this?

John Lewis Highsmith| Profile March 20th, 2011 - 11:45 pmMashplum

I used a spreadsheet, too. I knew the value had to be the square of some number and consist of 4 digits. The value of the square has to be between 1000 and 9999, which puts the numbers themselves between 32 and 99.

I entered 32 in A1 and in B1 put =A1^2. In A2 I entered =A1+1 and in B2 =A2+1 The reason for that last move was it would allow me to highlight both columns simultaneously, from the 2 row, down to 68. Next was a “Fill Down” and then check for the value that met the criteria. I did not do a good job of checking, because I missed the 2025 value.

bobble611| Profile March 21st, 2011 - 5:09 pm2025, 3025 and 9081

RK| Founder | Profile March 22nd, 2011 - 11:05 amHi Mashplum- Bilbao is unable to visit for a couple reasons, but he emailed me back and said ‘no’ in answer to your question.

(not that he was aware of)

Thanks

mshemim| Profile March 29th, 2011 - 12:51 pm9801

98+01=99

99*99=9801 is the answer

Bobo The Bear| PUZZLE MASTER | Profile March 30th, 2011 - 7:27 pm@Shawn: For your 8-digit numbers, are you requiring that all 8 digits be different (in which case, I could only find one), or merely that it breaks down into four separate, but different, 2-digit numbers (in which case, I can find four, or six if you allow for leading zeroes – a common occurrence in ticket numbers)?

RMW| Profile April 1st, 2011 - 8:41 am2025, 3025, and 9801 are the only three four-digit numbers having this property. Here’s some quick MATLAB/Octave code to confirm it:

%%%

test = 1111:9999;

truth = zeros(length(test));

for n = 1:length(test)

hundreds = fix(test(n)/100);

rest = test(n)-hundreds*100;

truth(n) = ((hundreds+rest)^2 == test(n));

end

index = find(truth);

winner = test(index)

%%%

Hex| PUZZLE MASTER | Profile April 3rd, 2011 - 4:50 am2 minutes to write a program, less than a second to output the result:

9801

Thanks Bilbao

vishakr| Profile December 5th, 2011 - 6:24 am1015;2025