Problem of the Day XVI
February 17th, 2015 at 3:19:22 PM
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Let n be a four-digit number whose square root is three times the sum of the digits of n. What is n ?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 17th, 2015 at 3:28:58 PM
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Quote: WizardLet n be a four-digit number whose square root is three times the sum of the digits of n. What is n ?
0.000?
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 17th, 2015 at 3:36:04 PM
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The square root is my age....
February 17th, 2015 at 3:57:56 PM
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Quote: SOOPOOThe square root is my age....
2,916, which has a square root of 54 and the sum of its digits is 18.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
February 17th, 2015 at 4:09:57 PM
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Quote: PlayYourCardsRight2,916, which has a square root of 54 and the sum of its digits is 18.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
I agree. I used a process of elimination too.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 17th, 2015 at 4:49:08 PM
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Quote: WizardQuote: PlayYourCardsRight2,916, which has a square root of 54 and the sum of its digits is 18.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
I agree. I used a process of elimination too.
Is there a strictly mathematical way (like a formula) to solve the problem? Or a list of equations? I think there is.
Where abcd ends up being 2916.
n = a*1000 + b*100 + c*10 + d
3*(a + b + c + d) = n^0.5
(a*1000+b*100+c*10+d) = (3*(a+b+c+d))^2
Is that right?
Would you still have to "plug n chug" (guess/eliminate)?
February 17th, 2015 at 4:59:25 PM
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Quote: RS(a*1000+b*100+c*10+d) = (3*(a+b+c+d))^2
Is that right?
Would you still have to "plug n chug" (guess/eliminate)?
Yes. With one equation and four unknowns, there would probably be an infinite number of solutions, if not confined to the positive integers. I think trial and error is pretty much the only practical way. Maybe MustangSally will prove me wrong.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 17th, 2015 at 5:02:33 PM
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Quote: RSQuote: WizardQuote: PlayYourCardsRight2,916, which has a square root of 54 and the sum of its digits is 18.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
I agree. I used a process of elimination too.
Is there a strictly mathematical way (like a formula) to solve the problem? Or a list of equations? I think there is.
Where abcd ends up being 2916.
n = a*1000 + b*100 + c*10 + d
3*(a + b + c + d) = n^0.5
(a*1000+b*100+c*10+d) = (3*(a+b+c+d))^2
Is that right?
Would you still have to "plug n chug" (guess/eliminate)?
I was about to start writing out a system of equations, but then I looked at the clock and realized I should leave work and go home :).
"So drink gamble eat f***, because one day you will be dust." -ontariodealer
February 17th, 2015 at 5:29:38 PM
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Quote: AcesAndEightsQuote: RSQuote: WizardQuote: PlayYourCardsRight2,916, which has a square root of 54 and the sum of its digits is 18.
I knew to be a four-digit number, the square root of n had to be between 32 and 99.
To multiply the sum of the digits times 3, the square root of n had to be divisible by 3 ... 33, 36, 39, etc.
Then it was a matter of elimination.
I'm sure there's a better way of solving, but that's my method.
I agree. I used a process of elimination too.
Is there a strictly mathematical way (like a formula) to solve the problem? Or a list of equations? I think there is.
Where abcd ends up being 2916.
n = a*1000 + b*100 + c*10 + d
3*(a + b + c + d) = n^0.5
(a*1000+b*100+c*10+d) = (3*(a+b+c+d))^2
Is that right?
Would you still have to "plug n chug" (guess/eliminate)?
I was about to start writing out a system of equations, but then I looked at the clock and realized I should leave work and go home :).
Me too, except I had to leave home and go to work. :(
February 17th, 2015 at 5:40:16 PM
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Let N = a*1000 + b*100 + c*10 + d.Quote: RS(a*1000+b*100+c*10+d) = (3*(a+b+c+d))^2
Then, reducing both sides mod 9 (using the fact that all powers of 10 are congruent to 1 mod 9), we get,
(a + b + c + d) ==9*(a+b+c+d)^2 == 0 (mod 9),
So, 9 divides a + b + c + d. Write a + b + c + d = 9*M.
Then N == 9*9^2*M^2 == 0 mod 9^3.
So, 9^3 = 729 must divide evenly into the answer. This lessens the workload a bit.
N = 9^3*M^2. For N < 10,000 and N > 1000, the only possible values for M are 2 and 3. This reduced the problem to two numbers to check,
N = 9^3*4 and N = 9^5.
The answer is N = 9^3*4.
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February 17th, 2015 at 5:43:59 PM
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Bravo to those who devised an answer without resorting to a spreadsheet. The raw brain horsepower on this board is impressive... even a little scary.
Simplicity is the ultimate sophistication - Leonardo da Vinci
February 17th, 2015 at 7:03:26 PM
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Quote: teliot...
N = 9^3*M^2. For N < 10,000 and N > 1000, the only possible values for M are 2 and 3. This reduced the problem to two numbers to check,
N = 9^3*4 and N = 9^5.
The answer is N = 9^3*4.
Is that really "N = 9^5" or possibly "N = 9^4" ?
I've never really studied integer mathematics, so I would never have come up with the idea of a mod 9 solution. Well done.
February 17th, 2015 at 7:55:10 PM
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I showed that N = 729*(M^2), with M = 2 or 3 as possible solutions. Thanks for the correction, 9^4.Quote: DocIs that really "N = 9^5" or possibly "N = 9^4" ?
Here's a similar one. In the new testament, Peter caught 153 fish. Why 153?
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February 17th, 2015 at 8:04:56 PM
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Why can't you let n be a one digit number, its easier to guess that way.Quote: WizardLet n be a four-digit number?
February 17th, 2015 at 8:06:49 PM
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Quote: teliotWhy 153?
I think that's been a topic of debate and disagreement for centuries.
February 17th, 2015 at 8:32:36 PM
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What three digit numbers equal the sum of the cubes of their digits? That's a whole lot of Holy Trinity for a number, don't you think?Quote: DocI think that's been a topic of debate and disagreement for centuries.
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February 17th, 2015 at 9:56:02 PM
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Quote: Doc
I've never really studied integer mathematics, so I would never have come up with the idea of a mod 9 solution. Well done.
Indeed. Nice work Eliot.
February 17th, 2015 at 11:18:25 PM
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As I went about this b/4 looking at spoilers, I figured a Range of 32^2 to 99^2. Since Sigma (d) is multiplied by 3, I also started to test the multiples of 3 starting with 33. By the time I got to 39, it became evident that there was a pattern involving sigma (d). Thus I tested the correct answer, and it agreed.
Extra Point: what was the pattern of sigma (d)?
Extra Point: what was the pattern of sigma (d)?
Some people need to reimagine their thinking.
February 18th, 2015 at 3:36:53 AM
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Quote: teliotHere's a similar one. In the new testament, Peter caught 153 fish. Why 153?
153 = 1^3 + 5^3 +3^3. This makes 153 a "narcissistic number." Maybe it is suggesting Peter was a narcissist.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
February 18th, 2015 at 3:53:16 AM
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Quote: teliotHere's a similar one. In the new testament, Peter caught 153 fish. Why 153?
1+2+...+16+17
February 18th, 2015 at 11:52:39 AM
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Quote: WizardLet n be a four-digit number whose square root is three times the sum of the digits of n. What is n ?
Resorting to the general or first principles of the divisibility of integers, the square root here is divisible by both three and nine, ie, 27. The sum here is divisible by nine because the number is.
( http://en.wikipedia.org/wiki/Divisibility_rule )
The square root here in the acceptable range is 54 or 81.
Quote: AyecarumbaThe raw brain horsepower on this board is impressive... even a little scary.
Old as the hills.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
February 18th, 2015 at 12:37:36 PM
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Read my post.Quote: KerkebetResorting to the general or first principles of the divisibility of integers, the square root here is divisible by both three and nine, ie, 27.
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February 18th, 2015 at 1:31:30 PM
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Thanks to teliot - I think I understand the logic and how to get to 0 mod 27.
Square = abcd = (1000*a+100*b+10*c+d) = (999+1)a+(99+1)b+(9+1)c+d = [999a+99b+9c]+[a+b+c+d].
Root = 3*(a+b+c+d) so the Square, being (3y)^2 = (9 y^2), is divisible by 9 (or 0 mod 9)
Therefore [999a+99b+9c]+[a+b+c+d] = 0 mod 9
Therefore [a+b+c+d] = 0 mod 9 (i.e. is 9 18 or 27, since each of a,b,c,d is [0-9] and 0000 9999 aren't valid squares).
But Root = 3*(a+b+c+d) (i.e. is 27 54 81), so is 0 mod 27 .
27 is too small to have four digit square, leaving 54 or 81.
Square = abcd = (1000*a+100*b+10*c+d) = (999+1)a+(99+1)b+(9+1)c+d = [999a+99b+9c]+[a+b+c+d].
Root = 3*(a+b+c+d) so the Square, being (3y)^2 = (9 y^2), is divisible by 9 (or 0 mod 9)
Therefore [999a+99b+9c]+[a+b+c+d] = 0 mod 9
Therefore [a+b+c+d] = 0 mod 9 (i.e. is 9 18 or 27, since each of a,b,c,d is [0-9] and 0000 9999 aren't valid squares).
But Root = 3*(a+b+c+d) (i.e. is 27 54 81), so is 0 mod 27 .
27 is too small to have four digit square, leaving 54 or 81.
February 18th, 2015 at 1:57:39 PM
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I showed analytically it is 0 mod 729. In fact I showed that the answer is 729×(M^2)
What did I do wrong in presenting my argument that people seem not to understand this conclusion?
What did I do wrong in presenting my argument that people seem not to understand this conclusion?
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February 18th, 2015 at 2:21:47 PM
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Quote: teliotWhat did I do wrong in presenting my argument that people seem not to understand this conclusion?
Well, my initial reply was:
Quote: DocI've never really studied integer mathematics, so I would never have come up with the idea of a mod 9 solution.
I expect that statement applies to most members of this forum. I suspect a lot of us might even have a good bit of difficulty following your derivation/proof no matter how well you presented it. Most folks might be hesitant to admit that -- some might even try to claim they were the one that developed the mod 9 proof. I dunno.
February 18th, 2015 at 3:15:59 PM
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Firstly I had not managed to think of your elegant solution, so thanks for adding the logical method rather than the guessing route I'd taken (after realising the 33,36,39 answers). I did not intend to imply you'd done anything wrong, but because I'm a bit rusty with my maths, couldn't see how the extra 3 (proving that the root was 0 mod 27 rather than 0 mod 9) had come from without working it through. I suspect that some things that were obvious to you, i.e. you could go from A to B, weren't to me, so I had to go slightly slower. While your "729*M^2" for the square is correct, I found it easier to think of the smaller value series that the root is 27,54,81... Having done that I thought I'd share the logic of my understanding of your solution.Quote: teliotWhat did I do wrong...
February 18th, 2015 at 3:36:52 PM
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Here is a similar question (thanks to Michael for the idea of asking this question), that follows the same type of derivation as I gave above.
Show that if N > 0 is such that the cube root of N is equal to 2 times the sum of its digits, then N = (5832)*(K^3), for some K > 0. There is no restriction here on the size of N.
Use this formula for N to find a solution to the question.
Then, if you are really eager (because this is the hard part), show that this solution is unique. That is, there is exactly one positive integer N > 0 whose cube root is 2 times the sum of its digits.
In coming up with this question, I wrote a computer program that let me vary the problem to be: "find an integer N whose x-th root is y-times the sum of its digits" and searched by brute force for a solution. Mike gave x = 2 and y = 3. I only found this additional case, with x = 3 and y = 2. It would be interesting to see if there are other (x,y) pairs with a non-trivial solution.
Show that if N > 0 is such that the cube root of N is equal to 2 times the sum of its digits, then N = (5832)*(K^3), for some K > 0. There is no restriction here on the size of N.
Use this formula for N to find a solution to the question.
Then, if you are really eager (because this is the hard part), show that this solution is unique. That is, there is exactly one positive integer N > 0 whose cube root is 2 times the sum of its digits.
In coming up with this question, I wrote a computer program that let me vary the problem to be: "find an integer N whose x-th root is y-times the sum of its digits" and searched by brute force for a solution. Mike gave x = 2 and y = 3. I only found this additional case, with x = 3 and y = 2. It would be interesting to see if there are other (x,y) pairs with a non-trivial solution.
I did the "unique" part by showing that any solution (if it exists) must be less than 10000000. I then just checked those possibilities by hand using the formula for N above, namely, N = (5832)*(K^3). You only need to check the values of K up to K = 11, because after that N > 10000000.
Here it is, written out by hand -- in a bit too much detail:
http://jacobsongaming.com/img/ForPosting/FunProblem_02_18_15.pdf
Here it is, written out by hand -- in a bit too much detail:
http://jacobsongaming.com/img/ForPosting/FunProblem_02_18_15.pdf
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February 18th, 2015 at 4:53:47 PM
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Thanks, Eliot, for posting. It took me some time to absorb what you were saying, but now I have a partial grasp on it.
As for your new problem:
part 1:
(10^3 d + 10^2 c + 10 b + a)^(1/3) = 2*(d + c + b + a)
(10^3 d + 10^2 c + 10 b + a) = 8*(d + c + b + a)^3
Reducing by mod 9, d + c + b + a == 0 mod 9
0 mod 9 == 8*(0 mod 9)^3 == 8*(0 mod 729) == 0 mod 5832
The only solution is 157464.
part 2:
There quickly becomes a point where the length of a digit * 9 (the maximum sum of the digits) is exceeded by N^(1/3)/2. For instance, the minimum cube root for any 8 digit N = 215.44. 215.44/2 > 8*9.
As for your new problem:
part 1:
(10^3 d + 10^2 c + 10 b + a)^(1/3) = 2*(d + c + b + a)
(10^3 d + 10^2 c + 10 b + a) = 8*(d + c + b + a)^3
Reducing by mod 9, d + c + b + a == 0 mod 9
0 mod 9 == 8*(0 mod 9)^3 == 8*(0 mod 729) == 0 mod 5832
The only solution is 157464.
part 2:
There quickly becomes a point where the length of a digit * 9 (the maximum sum of the digits) is exceeded by N^(1/3)/2. For instance, the minimum cube root for any 8 digit N = 215.44. 215.44/2 > 8*9.
I heart Crystal Math.
February 18th, 2015 at 4:54:14 PM
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Before you put the spoiler there are two parts.
The first uses a similar analysis based on mod 9.
Part One - Show Cube=5832*x^3.
Say the cube is 100000*a + 10000*b + 1000*c + 100*d + 10*e + f, then as above it is 9*something + [a+b+c+d+e+f] {and similarly for any number of digits in the cube}.
The root = 2 * [a+b+c+d+e+f]. Thus the cube is also 8* [a+b+c+d+e+f]^3.
Consider different values (mod 9)
Thus [a+b+c+d+e+f] has to be 0 mod 9.
Thus the root has to be 0 mod 9 and even, i.e. 0 mod 18. (0, 18,36, ...)
Thus the cube has to be 0 5832 46656 (i.e. 5832*x^3).
Part Two Prove 54 157464 is the unique answer.
Brute force can show the only answer for roots less than 200 is 54.
A cube of nnn has between 7 and 9 digits, hence its digit total cannot exceed 9*9=81, and twice that value not exceed 162.
A cube of nnnn has between 10 and 12 digits, hence its digit total cannot exceed 12*9=108, and twice that value not exceed 216 (i.e. has no chance to be a four digit number).
Any larger cubes suffer even worse!
The first uses a similar analysis based on mod 9.
Part One - Show Cube=5832*x^3.
Say the cube is 100000*a + 10000*b + 1000*c + 100*d + 10*e + f, then as above it is 9*something + [a+b+c+d+e+f] {and similarly for any number of digits in the cube}.
The root = 2 * [a+b+c+d+e+f]. Thus the cube is also 8* [a+b+c+d+e+f]^3.
Consider different values (mod 9)
[a...f] [a...f]^3 8[a...f]^3
0 0 0 (9 729 1458)
1 1 8 (1 1 8)
2 8 1 (2 8 64)
3 0 0 (3 27 54)
4 1 8 (4 64 512)
etc.
Thus [a+b+c+d+e+f] has to be 0 mod 9.
Thus the root has to be 0 mod 9 and even, i.e. 0 mod 18. (0, 18,36, ...)
Thus the cube has to be 0 5832 46656 (i.e. 5832*x^3).
Part Two Prove 54 157464 is the unique answer.
Brute force can show the only answer for roots less than 200 is 54.
A cube of nnn has between 7 and 9 digits, hence its digit total cannot exceed 9*9=81, and twice that value not exceed 162.
A cube of nnnn has between 10 and 12 digits, hence its digit total cannot exceed 12*9=108, and twice that value not exceed 216 (i.e. has no chance to be a four digit number).
Any larger cubes suffer even worse!
February 19th, 2015 at 3:49:00 PM
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Generalizing Mike's question: For any ordered pair (x,y) of integers with x >= 2, y >= 2, find all integers N such that the x-th root of N equals to y times the sum of the digits.
For example, for the pair (4, 3), the following integers have their 4-th root equal to 3 times the sum of their digits:
43046721
688747536
The following gives (x,y) pairs with 2 <= x, y, <= 10 together with any solutions N <= 10^18
For example, for the pair (4, 3), the following integers have their 4-th root equal to 3 times the sum of their digits:
43046721
688747536
The following gives (x,y) pairs with 2 <= x, y, <= 10 together with any solutions N <= 10^18
( 2, 2) 324
( 2, 2) 1296
( 2, 3) 2916
( 2, 4) 2704
( 2, 4) 5184
( 2, 4) 7744
( 2, 5) 400
( 2, 5) 2025
( 2, 5) 4225
( 2, 6) 11664
( 2, 7) 2401
( 2, 7) 12544
( 2, 8) 6400
( 2, 8) 20736
( 2, 8) 46656
( 2, 9) 26244
( 2, 9) 59049
( 2, 10) 100
( 2, 10) 8100
( 3, 2) 157464
( 3, 4) 64000
( 3, 4) 314432
( 3, 4) 1124864
( 3, 4) 1259712
( 3, 4) 1404928
( 3, 5) 729000
( 3, 5) 2460375
( 3, 6) 4251528
( 3, 6) 10077696
( 3, 7) 343000
( 3, 7) 2000376
( 3, 10) 1000
( 3, 10) 512000
( 3, 10) 4913000
( 3, 10) 5832000
( 3, 10) 17576000
( 3, 10) 19683000
( 4, 2) 8503056
( 4, 3) 43046721
( 4, 3) 688747536
( 4, 5) 65610000
( 4, 5) 332150625
( 4, 5) 2562890625
( 4, 7) 4032758016
( 4, 7) 9845600625
( 4, 8) 14003408896
( 4, 8) 23612624896
( 4, 8) 34828517376
( 4, 8) 68719476736
( 4, 9) 55788550416
( 4, 10) 10000
( 4, 10) 24010000
( 4, 10) 2342560000
( 4, 10) 3906250000
( 4, 10) 6146560000
( 4, 10) 16796160000
( 5, 2) 14693280768
( 5, 2) 31757969376
( 5, 3) 44840334375
( 5, 4) 3200000
( 5, 4) 188956800000
( 5, 4) 1488827973632
( 5, 4) 2771746976768
( 5, 6) 3570467226624
( 5, 7) 2470770901501
( 5, 7) 37294844329568
( 5, 7) 51716086897993
( 5, 8) 15045919506432
( 5, 9) 10896201253125
( 5, 9) 58602385427607
( 5, 10) 100000
( 5, 10) 1721036800000
( 5, 10) 5252187500000
( 5, 10) 6046617600000
( 5, 10) 20596297600000
( 6, 2) 64000000
( 6, 2) 9685390482496
( 6, 3) 6053445140625
( 6, 3) 101559956668416
( 6, 4) 256096265048064
( 6, 4) 570630428688384
( 6, 5) 531441000000
( 6, 5) 148035889000000
( 6, 5) 432510009765625
( 6, 6) 387420489000000
( 6, 6) 2917096519063104
( 6, 7) 16390160963076096
( 6, 7) 35766080886362176
( 6, 8) 2176782336000000
( 6, 8) 36520347436056576
( 6, 8) 39671359416303616
( 6, 10) 1000000
( 6, 10) 34012224000000
( 6, 10) 8303765625000000
( 6, 10) 24794911296000000
( 6, 10) 68719476736000000
( 7, 2) 10030613004288
( 7, 2) 47829690000000
( 7, 2) 504189521813376
( 7, 3) 19891027786401116
( 7, 4) 164341563462254592
( 7, 10) 10000000
( 7, 10) 6122200320000000
( 7, 10) 104603532030000000
( 7, 10) 275126141110000000
( 7, 10) 525233501440000000
( 8, 2) 39062500000000
( 8, 2) 474373168346071296
( 8, 10) 100000000
( 9, 2) 387420489000000000
( 9, 10) 1000000000
( 10, 10) 10000000000
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February 19th, 2015 at 4:08:03 PM
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Quote: teliot
Here's a similar one. In the new testament, Peter caught 153 fish. Why 153?
Actually, 153 is a remarkable number.
The largest prime factor of 153 is 17, and as the Wizard pointed out, 153 = 1+2+3+4+. . . +16+17.
But also, 153 is equal to the alternating sum and the differences of the squares of the integers from 1 to 17 ! Try it and see.
Also, pi (153) = pi (15) x 3!, where pi(x) is the prime counting function. Thus, the number of primes less than 153 is equal to the number of primes less than 15 x 3!
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
February 19th, 2015 at 4:16:26 PM
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Quote: gordonm888Also, pi (153) = pi (15) x 3!, where pi(x) is the prime counting function. Thus, the number of primes less than 153 is equal to the number of primes less than 15 x 3!
Pi(15) = 6.
Pi(153) = 36.
So,
Pi(153) = Pi(15)*3!
In addition to the above,
153 = 1! + 2! + 3! + 4! + 5!
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February 19th, 2015 at 4:42:38 PM
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Quote: teliotWhy 153?
Just to be sure I'm not misunderstanding the arguments here, are any of you really claiming that any of these mathematical coincidences, or maybe the combination of all of them, is the reason that the Gospel reported 153 as the number of fish that were caught? (As in "why" it was 153?)
I suspect that with reasonable effort you could come up with a list of mathematical oddities/coincidences for most any integer. I think a few years ago there was some Hollywood celeb going on talk shows promoting the magical/mystical properties of some fairly small integer that kept showing up in his life. He liked to cite a number of truly bizarre examples, like the sum (or some manipulation) of integers representing letters in people's names or show titles, etc.
BTW, here is a cached version of a web page that at least attributes the "significance" of 153 to some religious expressions in Hebrew. Don't think I will accept that as proof either.
February 19th, 2015 at 4:50:11 PM
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My original source on the number 153 was "The Incredible Dr. Matrix" by Martin Gardner, which is a book that everyone who likes numbers should own.Quote: DocI suspect that with reasonable effort you could come up with a list of mathematical oddities/coincidences for most any integer.
Another source I often refer to is "The Penguin Dictionary of Curious and Interesting Numbers" by David Wells, who states: "In the New Testament the net that Simon Peter dreew from the sea of Tiberias held 153 fishes. This was inevitable interpreted numerologically by the early Church fathers, especially St. Augstine. Everyone who likes numbers should own this book as well.
Forgive please the cut & paste, but here is the relevant section from Wikipedia:
Quote: Wikithing
The Miraculous catch of 153 fish by Duccio, 14th century, showing Jesus and the 7 fishing disciples (with Peter leaving the boat).
The Gospel of John (chapter 21:1–14) includes the narrative of the Miraculous catch of 153 fish as the third appearance of Jesus after his resurrection.[6]
The precision of the number of fish in this narrative has long been considered peculiar, and many scholars, throughout history, have argued that 153 has some deeper significance. Jerome, for example, wrote that Oppian's Halieutica listed 153 species of fish, although this could not have been the intended meaning of the Gospel writer because Oppian composed Halieutica after the Gospel text was written, and at any rate never gave a list of fish species that clearly adds up to 153.[7][8] It has also been noted that the Tetragrammaton occurs 153 times in the Book of Genesis.[9]
Augustine of Hippo argued that the significance lay in the fact that 153 is the sum of the first 17 integers (i.e. 153 is the 17th triangular number), with 17 representing the combination of divine grace (the 7 gifts of the Spirit) and law (the Ten Commandments).[10][11] Theologian D. A. Carson discusses this and other interpretations and concludes that "If the Evangelist has some symbolism in mind connected with the number 153, he has hidden it well,"[12] while other scholars note "No symbolic significance for the number of 153 fish in John 21:11 has received widespread support."[13]
Writers claiming a major role for Mary Magdalene have noted that in Greek isopsephy her epithet "η Μαγδαληνή" bears the number 8 + 40 + 1 + 3 + 4 + 1 + 30 + 8 + 50 + 8 = 153, thus, it is suggested, revealing her importance.[14] Similarly, the phrase "τὸ δίκτυον" (the net) used in the passage bears the number 1224 = 8 × 153,[14] as do some other phrases. The significance of this is unclear, given that Koine Greek provides a choice of several noun endings[15] with different isopsephy values.[16] The number 153 has also been related to the vesica piscis, with the claim that Archimedes used 153 as a "shorthand or abbreviation"[14] for the square root of 3 in his On the Measurement of the Circle. However, examination of that work[17] does not find the number 153 used in that way.[4]
Evagrius Ponticus referred to the catch of 153 fish, as well as to the mathematical properties of the number (153 = 100 + 28 + 25, with 100 a square number, 28 a triangular number and 25 a circular number) when describing his 153-chapter work on prayer.[18] Louis de Montfort, in his fifth method of saying the Rosary, connects the catch of 153 fish with the number of Hail Marys said (3 plus 15 sets of 10),[19] while St Paul's School in London was founded in 1512 by John Colet to teach 153 poor men's children, also in reference to the catch.[20]
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February 19th, 2015 at 4:57:58 PM
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I wonder how many numerical coincidences could be generated to show that "Wizard of Odds" is somehow equivalent to "Michael Shackleford" and prove that his developing/evolving into that identity was part of the grand divine plan of some higher power?
February 19th, 2015 at 5:30:20 PM
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Please, you go first. I think you may find that these "coincidences" are not so easy. The number 153 was chosen because of its numerological properties; the only question is "which" properties.Quote: DocI wonder how many numerical coincidences could be generated to show that "Wizard of Odds" is somehow equivalent to "Michael Shackleford" and prove that his developing/evolving into that identity was part of the grand divine plan of some higher power?
Maybe this can help you get started:
"W" is the 4-th letter from the end of the alphabet, "d" is the 4-th letter from the start of the alphabet.
"r" is the 9-th letter from the end of the alphabet, "i" is the 9-th letter from the start of the alphabet.
"z" is the last letter of the alphabet, "a" is the first letter of the alphabet.
1, 4 and 9 are the first three perfect squares.
"Wizard" therefore corresponds to the numerical palindrome "491194."
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February 19th, 2015 at 8:14:37 PM
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Quote: teliot
Maybe this can help you get started:
"W" is the 4-th letter from the end of the alphabet, "d" is the 4-th letter from the start of the alphabet.
"r" is the 9-th letter from the end of the alphabet, "i" is the 9-th letter from the start of the alphabet.
"z" is the last letter of the alphabet, "a" is the first letter of the alphabet.
1, 4 and 9 are the first three perfect squares.
"Wizard" therefore corresponds to the numerical palindrome "491194."
That's actually really amazing. Particularly from someone whose screen-name, when spelled backward, is toilet, lol.
(no offense meant, by the way. I'm brand-new to this forum -hello everyone! - but I am enjoying it, including this thread.)
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
February 19th, 2015 at 8:22:39 PM
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Which is obviously (to me, anyway) why I chose that as my moniker here, instead of my actual name. To go one small step further, the following is a palindrome:Quote: gordonm888That's actually really amazing. Particularly from someone whose screen-name, when spelled backward, is toilet
T. Eliot, top bard, notes putrid tang emanating, is sad. I'd assign it a name: "gnat dirt upset on drab pot toilet."
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February 19th, 2015 at 10:11:50 PM
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Quote: teliotWhich is obviously (to me, anyway) why I chose that as my moniker here, instead of my actual name. To go one small step further, the following is a palindrome:
T. Eliot, top bard, notes putrid tang emanating, is sad. I'd assign it a name: "gnat dirt upset on drab pot toilet."
OK, now I'm laughing. Yours sincerely, arabrab (always seemed like it should be the name of a villain in LOTR to me).
If the House lost every hand, they wouldn't deal the game.
February 21st, 2015 at 8:27:30 AM
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Quote: teliotRead my post.
Go on to rule out the extraneous solutions without the proofs; I'll go back and READ it.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
February 21st, 2015 at 8:59:58 AM
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Quote: KerkebetGo on to rule out the extraneous solutions without the proofs; I'll go back and READ it.
A good decision is based on knowledge and not on numbers.
-- Plato
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February 24th, 2015 at 7:12:31 AM
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Quote: teliotA good decision is based on knowledge and not on numbers.
-- Plato
Couldn't do it, eh? Perhaps, too much room in the margins.
Reminds me of V's strangle hold on the arts: Dysfunctional seriousness. Half-baked. Others' trivia as one's own.
Understanding is important too.
I wasn't the one who called myself a wizard of anything, let alone of math [in the name of numbers of the bottom line], and hence couldn't analyze the simplest of math problems with fundamental knowledge of the integers; nor the one who solved it in selfish oblivion, and hence was compelled to exhaust what little elegance with brute force, simulated trivia. One strives to strike a balance. Avoid a "little knowledge" soon forgot, or over exploited in numerous wrong directions.
Fictitious Socrates sitting in a corner talking to himself about popularity contests? The application doesn't exactly correlate with the numbers, which work out whether we know or understand. What the population doesn't want or need is a bombardment of proofs at every turn, let alone of the self-evident. Should I care about the detailed physics of my cell phone, other than to avoid over exposure to the microwaves? Still free to use it with peace of mind on occasion, I hope.
If you can't have fun with it, no matter your personal intelligence, you need to give it up. Perhaps, the business of converting from math to gambling analogizes that of the archetypal boozer qua gambler on "a distinguished path" to religion. But, winding up where you started as you try to come anywhere.
I'll take a stab at your question of the fishes, even if you won't. Serious fun.
From what I recall, the ancient word ichthys unscrambles a lot of religious text. Somehow its letters equate to the number 153, and the fish-sketch used early on by the Christians as a secret symbol to conceal their activities among themselves. The fish have always represented God's battle for our souls. Hence, the place where good and evil meet; or more eloquently, where God confronts the Devil as part of God's being as all. Arising from oneself is both a good and bad thing, and puts you intimately on your counterpart. (Where the levels of infinity too become exhausted as sets to form pt and anti-pt, the "ghosts of the machine" become each other, and the machine, itself. Also, as the purely extraneous solutions are commanded to their corners.)
{Hexagonal [3+5+1] = 153} ≡ triangular [(3+5)+1+5+3].
Regular or non-reflective nine is the neutral sum of three and six, from 333 and 666. Seventeen is the neutral proper, reflective or anti- sum on the inclusive God's 3 or tri- side of the equation. The 1+5+3 reflects left from the One. Mathematically, three squared plus the Devil's confounding then of two cubed. God in Devil, versa; and each as intimately on the other.
Incidentally, 153/9 = 17, but doesn't remove the general extraneousness of 17 here.
What's the number 35153 to do with this? Nothing.
But the prime numbers do seem to propagate or resonate outward from some fundamental event. Like 7 partitioned into 3+1+3, by which also 313 is prime.
And so on... on our merry ways. What fun!
A new day.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
February 24th, 2015 at 8:01:22 AM
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This misquote is one of my pet peeves. Last weekend I was at a funeral. On the order of service, this "little knowledge" was also misquoted. The actual quote, from Alexander Pope, is,Quote: KerkebetAvoid a "little knowledge" soon forgot, or over exploited in numerous wrong directions.
A little learning is a dangerous thing.
Drink deep, or taste not the Pierian Spring;
There shallow draughts intoxicate the brain,
and drinking largely sobers us again.
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February 28th, 2015 at 10:44:47 AM
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Quote: teliotThis misquote is one of my pet peeves.
I suggest, you stop retreating to misquoting others (who misquoted one) and henceforth continuing to torture yourself with the whimsy.
"Twas well observed by my Lord Bacon, That a 'little knowledge' is apt to puff up, and make men giddy, but a greater share of it will set them right, and bring them to low and humble thoughts of themselves." (I put my part of this in quotes to separate it from the rest. Here, and before.)
http://www.phrases.org.uk/meanings/a-little-knowledge-is-a-dangerous-thing.html
Add-on: You gotta watch out for that 3*3*3 X 13 = 351.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
February 28th, 2015 at 4:03:21 PM
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Can someone knowledgeable in English grammar (Don?) please tell me what "tense" this is? The use of "ing" throughout makes it appear to be the "present continuous tense," but I am not sure.Quote: KerkebetI suggest, you stop retreating to misquoting others (who misquoted one) and henceforth continuing to torture yourself with the whimsy.
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February 28th, 2015 at 4:43:06 PM
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Quote: teliotCan someone knowledgeable in English grammar (Don?) please tell me what "tense" this is? The use of "ing" throughout makes it appear to be the "present continuous tense," but I am not sure.
Grammar is not my "expert subject," but I think that "I suggest" makes it present simple tense. Present continuous tense is, "I am suggesting."
"Mistreating," "misquoting," and "continuing" are used as gerunds and not verbs here. Notice that they do not have forms of "to be" in front of them.
March 1st, 2015 at 10:37:03 AM
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Quote: ThatDonGuyGrammar is not my "expert subject".
Don't look at me.
Ultimately, and to begin with all along: the language - like everything else - becomes what YOU make of it.
To wit, the "Atheist's Unified Religion Approach"?
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
March 1st, 2015 at 12:28:13 PM
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Quote: teliotA good decision is based on knowledge and not on numbers.
-- Plato
One of my favorite quotes is:
"If you torture numbers enough they will confess to anything."
Not sure who said it, though.
So many better men, a few of them friends, are dead. And a thousand thousand slimy things live on, and so do I.
March 2nd, 2015 at 3:38:57 PM
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Quote: gordonm888"If you torture your data long enough they will confess to anything." - Ronald Coase
True of anything in which man plays role. And, especially, of what he says and others say he says.
But, isn't this more an ad hoc argument over who's permitted to do the torturing, and for what purpose? Motives.
Certainly, there's no harm in asking stuff like: If what we perceive to be the dimensions are fundamentally multiplicative in nature, then why is the 0! conspicuously absent from the relevant sum of series 1!+2!+3!+4!+5! = 153? And, while the majority of theoretical physicists argue over the question of 10 (=2X5) or 11 (=2X5+1) dimensions, depending on which basic theory invoked, it would be handy to come up with anything solid about the 0th or 1st, depending on how those are to be properly counted. (I get hung up on the stuff of going directly from 1 BC to AD 1.)
Also you might want to call also this a form of torture, but however YOU wish to interpret the numbers, the numbers, themselves, are what they are. For scientists such as Einstein, who believed strongly that the physics follows the math and its simplest concepts (to an extent), those small differences can be quite telling among a minefield of theoretical possibilities.
Perhaps, it more comes down to a matter of passion or direction in life.
As far as the "narcissism" goes, the mathematicians don't really care about that either. But they are as entitled to have some fun with the numbers as may the others. If the religious want to write the number 153 as its digits to the powers of 333, respectively, then "more power to them". Perhaps, some of them would have better answered what was specifically asked in this thread.
In any event, I don't think it then would have become such a matter of fumbling and judgment, let alone torture.
Nonsense is a very hard thing to keep up. Just ask the Wizard and company.
