Interesting Number: CR4 Challenge (09/08/09)

Yep - 298. That took less than five minutes in Excel

And we get until October 15 to work on it before the answer is posted.

290 is not only a nontotient and a noncototient, it is also an untouchable (Source). I mention this because when you type 298 into wikipedia you are redirected to the page for 290, which it turns out, is a much more interesting number.

Hi,

to be more elaborate,

XYZ be the numbers of three digit.

Query: (X)²⁺(Y)²⁺(Z)² = XYZ/2

Apply x=2, Y=9, Z=8 in above equ,

then, (2)²⁺(9)²⁺(8)² = 149 (≈ 298/2)

Also the "0" before any number(000, 050) is invalid (Not applicable for number with decimal)

Hari

.500

Droll - albeit it's not a solution (given that the sum of the squares of the digits is 25).
The GA vote demon-strates that you can fool some of the people most of the time.

If the challenge had not specified a unique answer, I would have given a few in other number bases. (There should of course be an infinity of solutions...)

That was actually what I thought of doing also, except I was gonna use hex. Yeah, I know the decimal doesn't carry over, but... Can you do it in binary?

I ran hex, octal and binary in Excel but didn't find any more "interesting" numbers. Hex and octal have several numbers that are close (Δ=1~2 decimal), and 101 & 111 in binary almost work (Δ=1/2 decimal), but none are exact.

I'll ignore the (universally trivial) 000 in what follows.

Answering your question, and anticipating some others:

Other than the cheat 010, binary has no answer. Similarly, ternary only has 011, quaternary 020, quinary has no solutions, senary 030, septenary 022, octal 012, 032, and 040, and nonary has no solutions.

We already covered decimal, which has 050 plus the single "proper" solution 268
Undenary has improper solutions 033, duodecimal 060.
Tredecimal has improper solutions 053 and 062, and a single proper solution 2_11_11
Quattuordecimal has 070, quindenary 044, hexadecimal 080, septendecimal 013, base18 024, 074, and 090, decennoval 055, vigenary 0_10_0, base21 035, base22 0_11_0, base 23 066 and 095, base24 0_12_0, base25 none, base26 0_13_0.
Base27 has 077 and 2_24_22
Base28 has 036, 0_11_6 and 0_14_0, base29 none.
Trigesimal has 014, 0_14_4, 0_15_0 and 2_21_28
Fyzimal has run out of patience, but you can see that proper solutions will be quite sparse...

N.B. "proper" solutions for the equivalent problem with more than 3 digits are not possible, as 3*(n-1)² < n³/2 for all n ≥ 2

If you write "50" as a three-digit number "050" then it too is a solution.

And if you write Zero as "000" then the trivial solution works too, because half of nothing is still nothing.

I see three numbers which meet the criteria, one of them "trivial" if you want to declare it so.

They are: 000, 050 and 298

And to be honest I wrote a short BASIC program to find them.

If you total the squares of the digits of a three-digit number, you get exactly half its value. Find the number.

298

298, I solved by putting numbers into 3 Cells in Excel and concatenating them to make the three digit number then I set another cell to half that value and a final cell to the sum of the squares. I then solved so that sum of the squares equaled the value of half of the concatenated cell. So, not that elegant but it worked.

hi...

are the answer is 000

i am not sure 100%

i hope its true i want to be .....

Regards

Razan Omar

I'm not sure I really liked the challenge question because it had a relatively easy solution but I do like the official answer. No need to know how to program a computer or make a spreadsheet. Just paper and pencil and a little insight.

Thanks,

Jim

Insight...
I may be missing something - if so please say:

But:
The obvious w<=4 constraint was placed rather later than it might have been,
The constraint should include* (50-2.xmax) <= X < 50, and X is even (that will further limit the search for square sums). (Note that the numbers presented exceeded this search range for X)
More significantly, it seems to miss the useful constraint on x:
base².xmax/2 < 2.(base - 1)² + xmax²
*For number base of 10, that means that xmax is 3.

For me the "killer" is that even using the additional constraints the search for acceptable squares appears to be no smaller than a structured search using the raw numbers - plus the numbers are rather larger
I think this may be seen if we try to program the solution. I give an abbreviated version of the "natural" search - on the basis that the altered search is
First find xmax
For z = 0 to (base-1) step 2
For x = 1 to xmax
trial = base^2*x + base*y + z
b=sqrt(trial - x^2 - z^2)
if b=INT(b) then print solution
next x
next z

P.S. I have since rechecked the value of xmax for arbitrary number bases (I previously used a value of 4, though I thought it should be 3). It is indeed the minimum of 3 and (base-1). That means that the maximum number of "trials" that is needed is 3.INT((base+1)/2). This can readily be laid out on a spread-sheet.

That is how I did my trials - albeit I used a slightly different test:
sqrt(base⁴/4-x²-z²+z) has to be an integer

Yes the number is 298, but the line above says z=9, I realise you meant to write z=8, a simple typo, so thought I'd let you know so yoiu can fix it for posterity! Happy Christmas!