Math Quiz Thread

[beero1000]

That works too. My route was more complicated.

I arrived at it by doing the Taylor series for \((x^n-b)^{(1/n)}\) with b= x^n. When you do that you end up with:

x - x*first term + x*second term - x*third term.... dividing through by x you get:

1 - first term + second term - third term +....

and since the Taylor series adds up to 0 (for x^n-x^n), and the first term is 1, the rest of the terms = -1. So I flopped the signs on the rest, and got 1.

If you want to make arriving at 1 complicated, use my route.

It happens.

Here's one: Let F_n denote the n-th Fibonacci number (i.e. F₁ = F₂ = 1 and F_n = F_n-1 + F_n-2) and find \(\sum_{n=1}^\infty \frac{F_n}{2^n}\). In general, what is \(\sum_{n=1}^\infty \frac{F_n}{k^n}\)?

Bonus points for determining convergence. Extra bonus points for not using the Binet formula.

 

[lpetrich]

I will set
\(S(k) = \sum_{n=1}^\infty \frac{F_n}{k^n}\)
Then, in preparation for using the recurrence relation to subtract out the F's,
\( k S(k) = \sum_{n=1}^\infty \frac{F_n}{k^{n-1}} = 1 + \sum_{n=1}^\infty \frac{F_{n+1}}{k^n} \)
and
\( k^2 S(k) = \sum_{n=1}^\infty \frac{F_n}{k^{n-2}} = k + 1 + \sum_{n=1}^\infty \frac{F_{n+2}}{k^n} \)
Useing that recurrence relation,
\( (k^2 - k - 1) S(k) = k \)
or
\( S(k) = \frac{k}{k^2 - k - 1} \)
Thus,
S(2) = 2


This method can be used for *any* series given by a constant-coefficient recurrence relation.

 

[steve_bnk]

Take a look over here.

http://www.talkrational.org/forumdisplay.php?f=78

ipetrich somebody is talking about you...

http://www.talkrational.org/showthread.php?t=64468

 

[lpetrich]

I have better things to do than to go to that site.

That aside, here's a site that's full of logic and math puzzles: Jason Rosenhouse's Problem of the Week, a part of his math classes.

I'll give his two most recent ones. Substitute distinct digits for letters to make base-10 numbers with the appropriate arithmetic operations:

  S E N D
+ M O R E
---------
M O N E Y


  D O U B L E
  D O U B L E
+     T O I L
-------------
T R O U B L E

 

[beero1000]

I have better things to do than to go to that site.

That aside, here's a site that's full of logic and math puzzles: Jason Rosenhouse's Problem of the Week, a part of his math classes.

I'll give his two most recent ones. Substitute distinct digits for letters to make base-10 numbers with the appropriate arithmetic operations:

  S E N D
+ M O R E
---------
M O N E Y


  D O U B L E
  D O U B L E
+     T O I L
-------------
T R O U B L E

There's also the related

  S E N D
+ M O R E
+ G O L D
---------
M O N E Y

There are multiple letter allocations for the letters here but, oddly enough, the value of MONEY remains constant.

 

[lpetrich]

I found 12 solutions of SEND MORE GOLD MONEY:

2894   1058   7034   10986
7894   1058   2034   10986
2894   1038   7054   10986
7894   1038   2054   10986
4673   1586   9503   15762
9673   1586   4503   15762
4673   1506   9583   15762
9673   1506   4583   15762
4237   1582   9507   15326
9237   1582   4507   15326
4237   1502   9587   15326
9237   1502   4587   15326

SEND MORE MONEY:

9567   1085   10652

DOUBLE DOUBLE TOIL TROUBLE:

798064   798064   1936   1598064

 

[steve_bnk]

I have better things to do than to go to that site.

That aside, here's a site that's full of logic and math puzzles: Jason Rosenhouse's Problem of the Week, a part of his math classes.

I'll give his two most recent ones. Substitute distinct digits for letters to make base-10 numbers with the appropriate arithmetic operations:

  S E N D
+ M O R E
---------
M O N E Y


  D O U B L E
  D O U B L E
+     T O I L
-------------
T R O U B L E

I have real puzzles to solve, I don't need to create ones.

 

[lpetrich]

I wrote a program that does SEND MORE MONEY puzzles by brute force. I had to write it in C++ to get it to run fast, something that required some annoying bookkeeping. But I wrote it, and it takes less than a second to check every possibility, even though it generates all 10! = 3628800 possible digit permutations.

I verified SEND MORE MONEY and DOUBLE DOUBLE TOIL TROUBLE, but for SEND MORE GOLD MONEY I found 28 solutions:

9673 1586 4503 15762 
9673 1506 4583 15762 
4673 1586 9503 15762 
4673 1506 9583 15762 
9054 1730 6724 17508 
9054 1720 6734 17508 
6054 1730 9724 17508 
6054 1720 9734 17508 
7894 1058 2034 10986 
7894 1038 2054 10986 
2894 1058 7034 10986 
2894 1038 7054 10986 
9856 1378 2346 13580 
9856 1348 2376 13580 
7856 1398 4326 13580 
7856 1328 4396 13580 
4856 1398 7326 13580 
4856 1328 7396 13580 
2856 1378 9346 13580 
2856 1348 9376 13580 
9237 1582 4507 15326 
9237 1502 4587 15326 
4237 1582 9507 15326 
4237 1502 9587 15326 
9478 1634 5628 16740 
9478 1624 5638 16740 
5478 1634 9628 16740 
5478 1624 9638 16740

 

[lpetrich]

Jason Rosenhouse's Problem of the Week: Knights, Knaves, Normals, Werewolves and Other Fanciful Creatures In this genre of puzzles, knights always tell the truth and knaves always lie, and you have to use their statements to puzzle out who's who. Here's the first one:

You meet three people, whose names are Asimov, Buffy and Columbo. Asimov says, “All of us are knaves.” Buffy then says, “Actually, exactly one of us is a knight.” Which of the three are knights and which are knaves?

 

[Kharakov]

One statement has to be true, unless we apply a certain individual's standard of logic. Not allowed to guess their name!

 

[lpetrich]

The solution:

If Asimov is a knight, then he must be a knave, so he cannot be a knight. But if he is a knave, then his statement can be false with him still being a knave. Thus, there is at least one knight, but Asimov is not one of them.

If Buffy is a knight, then she is the only knight, and Columbo must be a knave. But if Buffy is a knave, then there must be two knights, meaning that she must be a knight.

Thus,

Asimov	knave
Buffy	knight
Columbo	knave

 

[lpetrich]

Here's another one. From Problem of the Week: Sam Loyd Tribute, here's the fifth one:

The left side balances the right side.

left side	right side
bottle, glass	pitcher
bottle	plate, glass
2 pitchers	3 plates
bottle	how many glasses?

 

[lpetrich]

The solution:

(bottle) = (plate) + (glass)
(pitcher) = (bottle) + (glass) = (plate) + 2(glass)
2(pitcher) = 2(plate) + 4(glass) = 3(plate)
(plate) = 4(glass)
(bottle) = 5(glass)
(pitcher) = 6(glass)

Thus, 1 bottle = 5 glasses.

You can find more at
Sam Loyd Official Site - Get off the Earth, Trick Donkeys, Pony Puzzle, Cyclopedia of Puzzles
Sam Loyd Puzzles
Sam Loyd's Cyclopedia of Puzzles
Puzzles by Sam Loyd | Thinks.com

Here's another one of his puzzles. Take the numbers 1 2 3 4 5 6 7 8 9 and arrange them in a 3*3 array, so that all 3 rows, all 3 columns, and all 2 diagonals add up to the same number. How many possible arrangements of these numbers satisfy those conditions? If you count as one arrangement all arrangements related by rotations and reflections, how many possible arrangements are there then?

The two questions are mine, however.

 

[lpetrich]

The solution to that number puzzle.

First, each row must add up to a number S that is the same for all rows. All three rows thus add up to 3S. Since each number is represented exactly once in that sum, 3S = 45 or S = 15. The same reasoning applies for the columns.

Now for the center member. Add up all the rows, columns, and diagonals that pass through the center. The sum of the numbers is (sum of off center) + 4*(center) = 4*15 = 60
Subtracting out the sum of all gives
3*(center) = 60 - 45 = 15
Thus, the center one is 5.
For each center row, column, and diagonal, the other two numbers must add up to 10:
1 5 9
2 5 8
3 5 7
4 5 6

Each number other than 5 must be on a row and/or a column. Let's see what possibilities there are.
1: 1 6 8
2: 2 4 9, 2 6 7
3: 3 4 8
4: 2 4 9, 3 4 8
6: 1 6 8, 2 6 7
7: 2 6 7
8: 1 6 8, 3 4 8
9: 2 4 9
So 2, 4, 6, 8 are on corners while 1, 3, 7, 9 are on sides.

One can go on a trip around the square with 2 9 . 4 3 . 8 1 . 6 7

All solutions will have this trip, starting at any corner and going in either direction. That gives a grand total of 8 solutions, and they are all related to any one of them by symmetry operations: rotations and reflections.

One of the solutions is
2 9 4
7 5 3
6 1 8


Number Puzzles has lots of them, like

Magic Square 111 Puzzle
Every row, column, and diagonal adds up to 111. You start with
X X 7
13 37 X
X X X

Algebra Puzzles has lot of "SEND + MORE = MONEY" puzzles:

ABCD * E = DCBA
ABC + DEF = GHIJ
ABCD * D = DCBA
ABC * DEF = 123456 if A = 1
CAT = (C + A + T) * C * A * T
COW * COW = DEDCOW
(DD)^E = DEED
C^L = LOGIC
(L+O+G+I+C)^3 = LOGIC
CHOO + CHOO = TRAIN
6 * KAYAK = SPORT

Now for some more Sam Loyd puzzles.

Outwitting the Weighing Machine Puzzle In it, 5 schoolgirls get weighed in pairs, so as to hide their weights. They get combined weights of 129, 125, 124, 123, 122, 121, 120, 118, 116, and 114 lbs. How much does each of them weigh individually?

 

[lpetrich]

Magic Square 111:
X X 7
13 37 X
X X X

2nd row: X = 61
2nd diagonal: X = 67
1st column: X = 31
3rd column: X = 43
1st row: X = 73
3rd row: X = 1

Solution:
31 73 7
13 37 61
67 1 43

 

[lpetrich]

ABCD * E = DCBA
2178 * 4 = 8712
(found with brute force)

ABC + DEF = GHIJ
{{246,789,1035},{264,789,1053},{324,765,1089},{342,756,1098},{347,859,1206},{423,675,1098},{426,879,1305},{432,657,1089},{437,589,1026},{473,589,1062},{624,879,1503},{743,859,1602}}
(found with brute force, with constraint A<D, B<E, C<F)

ABCD * D = DCBA
1089 * 9 = 9801
(found with brute force)

ABC * DEF = 123456 if A = 1
Solution:
Factor the product:
123456 = 2^6 * 3 * 643
Since 643 is a prime, then one of the multiplicands must be it. The other multiplicand is from the remaining factors.
Thus, the solution is
192 * 643 = 123456

CAT = (C + A + T) * C * A * T
135 = (1+3+5) * 1 * 3 * 5
(found with brute force)

COW * COW = DEDCOW
(no solution)
But turning the second one into DEFCOW, I find this solution:
625^2 = 390625
(found with brute force)

(DD)^E = DEED
Solution:
DD >= 11. To get a 4-digit number, E <= 3. So E = 2 or 3.
For E = 2, D = 1, 5, 6 -- no solution
For E = 3, D = 1, 4, 5, 6 -- all but D = 1 make a too-big number
Thus,
(11)^3 = 1331

C^L = LOGIC
Solution:
All these have 5 digits: 3^9, 4^7, 4^8, 5^6, 5^7, 6^6, 7^5, 8^5, 9^5
First-digit condition: 5^9, 9^5
Both of them satisfy the last-digit condition.
All digits different: 5^9 = 78125

(L+O+G+I+C)^3 = LOGIC
(1+9+6+8+3)^3 = 19683
(found with brute force)

CHOO + CHOO = TRAIN
2*5466 = 10932
2*5488 = 10976
(found with brute force)

6 * KAYAK = SPORT
6 * 15451 - 92706
(found with brute force)

I must be getting old and feeble I used Mathematica to find most of the solutions in brute-force fashion.

 

[lpetrich]

I did those SEND-MORE-MONEY problems with the help of these Mathematica functions:
Range -- to get the digits
Subsets -- to get all selections of some number of them
Permutations -- to get all possible orders of a seleciton
FromDigits -- to get a number from its digit values

One can also decompose a number into its digits with IntegerDigits. But Mathematica has programming functionality to make it possible to implement all those functions in terms of a few simple ones, though doing so would not be very efficient. For example, one can do digits to numbers, and polynomials more generally, with Horner's rule using the function Fold.

Now the schoolgirl problem. Start off with weights of pairs of them: 129, 125, 124, 123, 122, 121, 120, 118, 116, and 114 lbs. I'll find the solution using a rather complicated algorithm. I'll find a polynomial for the weight values, a 5th-order polynomial that will have 5 solutions. The coefficients have values
(-1)^(5-m) * sum (product of m weight values)
for the term with order 5 - m.

I found this solution: {56, 58, 60, 64, 65}

The Mathematica code that I'd written to solve this problem:

(* All sums of some number m of n x's -- for problem data *)

xsum[indxs_] := Total[x /@ indxs]

xsumall[n_, m_] := xsum /@ Subsets[Range[n], {m}]

(* Sum of product of m distinct x's -- total n x's -- for polynomial \
coefficients *)

xterm[indxs_] := Times @@ (x /@ indxs)

xtmsum[n_, m_] := Total[xterm /@ Subsets[Range[n], {m}]]

xtmsmall[n_] := Table[xtmsum[n, m], {m, n}]

(* List of powers of some list, then totals of those lists *)

lstpwrs[lst_, n_] := NestList[lst*# &, lst, n - 1]

lstpwrtot[lst_, n_] := Total /@ lstpwrs[lst, n]

(* Translation of problem data into polynomial coefficients *)

lstptopolc[lst_, n_] := Module[{k, ipr, lpt},
  lpt = lstpwrtot[lst, n];
  Table[Total[
    Table[c[ipr]*(Times @@ (lpt[[#]] & /@ ipr)), {ipr, 
      IntegerPartitions[k]}]], {k, n}]
  ]

(* Sum of two from five, with verification *)

coefs52 = {c[{1}] -> 1/4, c[{1, 1}] -> 1/24, c[{2}] -> -1/6, 
   c[{1, 1, 1}] -> 1/96, c[{2, 1}] -> -1/8, c[{3}] -> 1/3, 
   c[{1, 1, 1, 1}] -> 1/2304, c[{2, 1, 1}] -> -1/288, c[{3, 1}] -> 0, 
   c[{2, 2}] -> -1/72, c[{4}] -> 1/12, c[{1, 1, 1, 1, 1}] -> -1/56320,
    c[{2, 1, 1, 1}] -> 7/12672, c[{3, 1, 1}] -> -5/1584, 
   c[{2, 2, 1}] -> -7/3168, c[{4, 1}] -> 7/528, c[{3, 2}] -> 1/198, 
   c[{5}] -> -1/55};

xtmsmall[5] - lstptopolc[xsumall[5, 2], 5] /. coefs52 // Expand

(* Polynomal evaluation -- Horner's rule *)

polevaltb[x_, c_] := Fold[x*#1 + #2 &, 0, c]

polevalbt[x_, c_] := polevaltb[x, Reverse[c]]

wts = {129, 125, 124, 123, 122, 121, 120, 118, 116, 114};

Length[wts]

(* The solution *)

polevaltb[-w, Prepend[lstptopolc[wts, 5] /. coefs52, 1]] // Factor

wtsol = w /. Solve[% == 0, w]

Sort[Total /@ Subsets[wtsol, {2}]] === Sort[wts]