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Crossnumber

cinelli
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Crossnumber

#125324

Postby cinelli » March 16th, 2018, 11:06 am

.
---- ---- ----
|A | |B |
| | | |
| | | |
---- ---- ----
| |XXXX| |
| |XXXX| |
| |XXXX| |
---- ---- ----
|C | | |
| | | |
| | | |
---- ---- ----

In this crossnumber the clues are as follows:
Across
A = X^Y
C = Y^(W-Z)
Down
A = (W-X)^Z - Z*W
B = W*Y*Z + W

Cinelli

Gengulphus
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Re: Crossnumber

#127974

Postby Gengulphus » March 25th, 2018, 7:04 pm

cinelli wrote:
.
---- ---- ----
|A | |B |
| | | |
| | | |
---- ---- ----
| |XXXX| |
| |XXXX| |
| |XXXX| |
---- ---- ----
|C | | |
| | | |
| | | |
---- ---- ----

In this crossnumber the clues are as follows:
Across
A = X^Y
C = Y^(W-Z)
Down
A = (W-X)^Z - Z*W
B = W*Y*Z + W

Not-quite-complete spoiler:

I'm not familiar with crossnumber conventions, but I'll assume they include one that the solutions to the clues are non-negative integers of the exact lengths (when written in decimal) indicated in the spaces for them, so they can be (and are) written using only the digits 0 to 9, without either minus signs or leading zeros.

And to get the not-quite-completeness out of the way: it is conceivable that there are values of X, Y, Z, W that include one or more non-integers, such that X^Y, Y^(W-Z), (W-X)^Z - Z*W and W*Y*Z + W all come out as integers. Certainly it's possible for at least one of them to come out as an integer even if all four of X, Y, Z and W are non-integers - for example, if X is the cube root of 123^2 = 15129, Y is 1.5 and Z and W are any other two non-integers, then X^Y = (CubeRoot(123^2))^(3/2) = CubeRoot((123^2)^(3/2)) = CubeRoot(123^(2*3/2)) = CubeRoot(123^3) = 123. And that can easily be extended to making Y^(W-Z) come out as an integer as well, e.g. by choosing W and Z to be any pair of non-integers such that W = Z + LOG(789)/LOG(1.5), as then Y^(W-Z) = 1.5^(LOG(789)/LOG(1.5)) = 789. That still allows Z to be chosen arbitrarily, and since (W-X)^Z - Z*W and W*Y*Z + W are continuous functions of that choice of Z, it seems highly likely to me that either of them can be made an integer by an appropriate choice of Z (the only way it might not be possible is if the range of the function concerned is entirely contained strictly inside the gap between two consecutive integers). Getting it to come out as an integer in the range 100 to 999 might be more difficult because a less stringent requirement on the range of the function concerned is needed to make it impossible, and getting it to come out as one with the correct digits to fit in with the across solutions being 123 and 789 might be more difficult still because an even less stringent requirement is needed to make it impossible, but it still seems to me to be quite likely to be possible. So I think it quite likely that a suitable choice of X, Y, Z, W exists such that at least three of them (and probably all four) are non-integers, and at least three of X^Y, Y^(W-Z), (W-X)^Z - Z*W and W*Y*Z + W are integers, with the right intersection properties between those three. Especially as I could clearly have chosen the across solutions to be any integers in the range 100 to 999 instead of 123 and 789, and the same argument would apply analogously - so I only need it to turn out to be possible for one of the 900^2 = 810,000 possible choices of across solutions, rather than necessarily for the specific choice of 123 and 789.

At that point, however, I've fixed all four of X, Y, Z, W, so it's a matter of luck whether the fourth one of X^Y, Y^(W-Z), (W-X)^Z - Z*W and W*Y*Z + W also turns out to be an integer, and it is probably overwhelmingly unlikely (or in mathematical parlance, "has probability zero"). But there is still an outstanding arbitrary choice in the above, namely that of Y = 1.5, and a similar argument involving using a freely-chosen value of Y > 1, setting X, Z and W to make X^Y, Y^(W-Z), and one of (W-X)^Z - Z*W and W*Y*Z + W be integers, and regarding the other one of them as a function of the chosen value of Y might well make it possible to make it too be an integer. But the functions concerned are pretty complicated and I'm not at all certain whether their ranges would end up making a suitable fourth integer solution possible... The complexity is such that I don't fancy tackling the possibility that one or more of X, Y, Z, W not being an integer could lead to a valid solution, especially not within the limits of TLF posts!

So this spoiler is incomplete in that it doesn't deal with that possibility, either by finding valid solutions that arise from sets of X, Y, Z, W that include one or more non-integers or by proving that there are no such solutions. Or equivalently, it assumes that the crossnumber includes an unstated requirement that all four of X, Y, Z and W are integers.

On to actually solving the crossnumber:

First, look at what Y can be. It cannot be negative, because that would make X^Y a non-integer for every possible X, nor can it be zero, since that would make X^Y be either 0 or 1 (the latter if X is also zero) amd neither is a 3-digit integer. It also cannot be 1, because that would make Y^(W-Z) be 1 whatever the values of W and Z, nor can it be greater than 9, since that would make X^Y be 0, 1 or a greater-than-3-digit integer. So Y lies in the range 2 to 9.

Given that, we can easily list the possible 3-digit values of Y^(W-Z):

For Y=2, they are 128 (with Z=W+7), 256 (with Z=W+8) and 512 (with Z=W+9)
For Y=3, they are 243 (with Z=W+5) and 729 (with Z=W+6)
For Y=4, it must be 256 (with Z=W+4)
For Y=5, they are 125 (with Z=W+3) and 625 (with Z=W+4)
For Y=6, it must be 216 (with Z=W+3)
For Y=7, it must be 343 (with Z=W+3)
For Y=8, it must be 512 (with Z=W+3)
For Y=9, it must be 729 (with Z=W+3)

For each of those, we know the value of Y is C and we know that W = Z+K for particular constants C and K. So W*Y*Z + W = (Z+K)*C*Z+Z+K = C*Z^2 + (C*K+1)*Z + K is a known quadratic function of Z, and is a 3-digit number that ends on the same digit as the known value of Y^(W-Z). Furthermore, we know that Z is not negative (if it were, (W-X)^Z - Z*W would not be an integer) and that C and K are positive, so all three of C*Z^2, (C*K+1)*Z and K are positive and the first two increase as Z increases, while K of course just remains constant. So C*Z^2 + (C*K+1)*Z + K increases as Z increases, which means that we can find the possible 3-digit values just by starting at Z=0 and increasing Z by 1 at a time, evaluating the quadratic at each value of Z, starting recording the values when they reach 3 digits long, and stopping the entire process when they reach 4 digits long. That's a bit tedious, but do-able - here are the results, with the 3-digit values that share their last digit with Y^(W-Z) marked with stars:

Y^(W-Z)  C  K  quadratic   3-digit values
----------------------------------------------------------------------------------------
128 2 7 2Z^2+15Z+7 132,169,210,255,304,357,414,475,540,609,682,759,840,925
256 2 8 2Z^2+17Z+8 108,143,182,225,272,323,378,437,500,567,638,713,792,875,962
512 2 9 2Z^2+19Z+9 117,154,195,240,289,*342*,399,460,525,594,667,744,825,910,999
243 3 5 3Z^2+16Z+5 117,160,209,264,325,392,465,544,629,720,817,920
729 3 6 3Z^2+19Z+6 130,176,228,286,350,420,496,578,666,760,860,966
256 4 4 4Z^2+17Z+4 *136*,189,250,319,*396*,481,574,675,784,901
125 5 3 5Z^2+16Z+3 147,208,279,360,451,552,663,784,*915*
625 5 4 5Z^2+21Z+4 112,168,234,310,396,492,598,714,840,976
216 6 3 6Z^2+19Z+3 114,175,248,333,430,539,660,793,938
343 7 3 7Z^2+22Z+3 132,*203*,288,387,500,627,768,*923*
512 8 3 8Z^2+25Z+3 150,231,328,441,570,715,876
729 9 3 9Z^2+28Z+3 168,*259*,368,495,640,803,984

So we have 7 possible combinations of Y^(W-Z) and W*Y*Z + W, one for each starred entry in the above table, and for each of them, we know the corresponding values of Y (= C in the above table), Z (= the value for which the quadratic in the above table is equal to the starred entry) and W (= Z + K in the above table). The next step is to determine the possible values of X: they are the values for which X^Y is a 3-digit number and ends with the digit that starts W*Y*Z + W. For Y=2, they are those of the 22 squares 10^2, 11^2, ..., 31^2 that end with the right digit; for Y=3, those of the 5 cubes 125, 216, 343, 512 and 729 that end with the right digit; for Y=4, those of the two fourth powers 256 and 625 that end with the right digit; for Y = 5, 6, 7, 8 and 9, they are 243, 729, 128, 256 and 512 respectively if they end with the right digit. Note that when Y is even, X can potentially be either positive or negative; when Y is odd, it has to be positive.

The results are:

Y   Z   W  Y^(W-Z)  W*Y*Z + W  Possible values of X
---------------------------------------------------
2 9 18 512 342 ---
4 4 8 256 136 ---
4 8 12 256 396 ---
5 12 15 125 915 ---
7 4 7 343 203 ---
7 10 13 343 923 ---
9 4 7 729 259 2

With only one possibility left, it has to produce the correct solution to the crossnumber, but we'd better check it just in case the puzzle is ill-posed:

X=2, Y=9, Z=4, W=7

produces:

Across
A = X^Y = 2^9 = 512
C = Y^(W-Z) = 9^(7-4) = 9^3 = 729
Down
A = (W-X)^Z - Z*W = (7-2)^4 - 4*7 = 5^4 - 28 = 625 - 28 = 597
B = W*Y*Z + W = 7*9*4 + 7 = 259
.
---- ---- ----
|A | |B |
| 5 | 1 | 2 |
| | | |
---- ---- ----
| |XXXX| |
| 9 |XXXX| 5 |
| |XXXX| |
---- ---- ----
|C | | |
| 7 | 2 | 9 |
| | | |
---- ---- ----


Gengulphus

cinelli
2 Lemon pips
Posts: 153
Joined: November 9th, 2016, 11:33 am
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Re: Crossnumber

#129154

Postby cinelli » March 31st, 2018, 11:54 am

Bravo, Gengulphus.

Cinelli


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