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

Posted September 07, 2009 5:01 PM

This week's Challenge Question:

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

Let's say that the three digits of the number are x, y, and z, so that the number is xyz. Now, according to the statement these digits should satisfy the following equation

From the above equation we see that z must be an even number. Let say that z = 2w. With this definition, te above equation becomes

Now, let's group the equal variables and complete the square to get

It clear that we must have w < 5.

Now redefine the parentheses in the above equation. Let X = 50 - 2x, Y = 2y – 5, and note that we must have X < 50, Y <= 13.

Now, let's assign the possible values to w and check if it satisfy the constraints.

a)

all these number are not acceptable.

Now try for w = 1, and the results are similar. Only when w = 4, we can satify the constraints:

b)

but Y <=13, then the solution is X = 46 and Y = 13

Therefore, x =2, y = 9 and z = 9.

The number is 298.

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#1

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

09/08/2009 9:54 AM

298. The end (unless/until someone can explain why this is interesting)?

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#2

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

09/08/2009 11:04 AM

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.

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#3

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

09/08/2009 4:11 PM

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.

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#4

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

09/08/2009 5:40 PM

hi Roger,

I had a quick look at your reference, and this seems to be an area where Wikipedia is highly fallible. For example, it states that five is untouchable and four is not untouchable. However, 5 (=2+3) is actually the sum of all the proper* divisors of 6; and 4 is untouchable, as the only additive decompositions are (3+1) and (2+2) neither of which can be the sum of proper divisors.

However, I wouldn't wish to comment on 290 - it's rather a long search

*Neither unity nor the number itself are proper divisors

P.S. I am not qualified to say whether it is the definition or the selection that is incorrect, though my suspicion would be the selection.

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#22

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

09/24/2009 8:51 AM

Hi Fyz,

Sorry it took me so long to respond. I was interested in your reply and did some research.

I think 5 is indeed an untouchable number (though I just learned this stuff so I may have made a mistake.

An untouchable number is a positive integer that cannot be expressed as the sum of all of the proper divisors of an positive integer.

You pointed out that 2+3=5, and 2 and 3 are proper divisors of 6, which is true, but 1 is also a proper divisor of 6, meaning that if you sum all the proper divisors of 6 you get 2+3+1=6.

So I think 5 is indeed an untouchable number.

In your defense, I had no idea what an untouchable number was in my original post, but I'm glad I do now.

Roger

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#23

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

09/26/2009 4:35 PM

Hi Roger

Thanks for pointing out the error of my ways. I hadn't realised that a "Proper Divisor" was different from a "Proper Factor". According to the definition you give (and I have now found elsewhere), 1 is a "proper divisor" and therefore 5 is indeed untouchable.

There is consistent logic to the definition of a proper factor - but I'm d----d if I can find it in my heart to admit the same of the definition of a Proper Divisor.

It makes me wonder whether there is an equivalent name for numbers that cannot be built from the sums of proper factors; however, it's probably not worth any trouble, as the only difference is that these are the Untouchable Numbers minus unity.

Regards

Fyz

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#10

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

09/09/2009 2:08 AM

Well Roger, you seem also to have found the answer to dac's question

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#9

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

09/09/2009 1:24 AM

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

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#11

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

09/09/2009 4:32 AM

There is a frivolous proof to show that all numbers are interesting. Separate all numbers into two sets; one interesting the other uninteresting. There will be a smallest number of the uninteresting set, which thereby acquires interest and gets removed, which leads to the next one and so on ad infinitum. Thereby rendering all numbers interesting.

Bioramani

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#12

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

09/09/2009 4:42 AM

Priceless, worth a point

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#5

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

09/08/2009 7:45 PM

.500

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#13

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

09/09/2009 4:50 AM

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...)

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#15

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

09/09/2009 7:29 AM

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?

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#17

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

09/09/2009 12:46 PM

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.

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#18

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

09/09/2009 1:31 PM

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

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

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#6

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

09/09/2009 12:09 AM

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.

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#7

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

09/09/2009 12:28 AM

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.

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#8

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

09/09/2009 12:33 AM

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

298

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#14

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

09/09/2009 6:26 AM

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.

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#16

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

09/09/2009 10:20 AM

hi...

i am not sure 100%

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

Regards

Razan Omar

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#19

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

09/16/2009 10:19 AM

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

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#20

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

09/17/2009 6:20 AM

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

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#21

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

09/18/2009 5:50 AM

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

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#24

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

12/22/2009 8:40 AM

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!

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