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:
The 12 Dots
Nine Brothers and Sisters
2025
2025
How about 2025?
(20 + 25)^2 = 45^2 = 2025
9801 works, too.
I also find only 2 cases where this would work for an 8-digit number…
9801
9801====>98+01=99
99 × 99 = 9801
To tell the truth,I found it in the net.
An 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!!
Answer = 9801
98 + 01 = 99
99 * 99 = 9801
Ok 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.
9801
I believe the answer is 2025 20+25=45 45*45=2025
Just my 2 cents worth
Steve
If 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.)
The value is 9801, which is the square of 99.
I think “9801″
ripped down the centre, you get 98 and 1 = 99, 99 squared is 9801
Good one!
2025
20+25=45
45*45=2025
9801
9801
98 + 1 = 99
99×99 = 9801
nice one Bilbao
98 and 01
98*01 = 99
99^2 = 9801
Hi there guys, glad to see you liked this one : )
The answer Bilbao had for this was 9801
@Bilbao
I used Excel to figure this out by trial and error.
Is there a simple/clever/elegant way to solve this?
Mashplum
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.
2025, 3025 and 9081
Hi 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
9801
98+01=99
99*99=9801 is the answer
@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)?
2025, 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)
%%%
2 minutes to write a program, less than a second to output the result:
9801
Thanks Bilbao
1015;2025