Re: Enigma 1446 - Double entendre
- From: Mark Tilford <ralphmerridew@xxxxxxxxx>
- Date: Tue, 31 Jul 2007 17:16:46 GMT
On 2007-07-31, Chappy <petergregorychapman@xxxxxxxxxxx> wrote:
Enigma 1446 - Double entendre
New Scientist magazine, 9 June 2007.
by Susan Denham.
For each answer in these cross-figures, two
clues have been given; one applies to each
figure, but in no particular order.
+-----+-----+-----+ +-----+-----+-----+
|1 : |2 | |1 : |2 |
| : | | | : | |
+-----+-----+-----+ +-----+-----+-----+
| |3 : | | |3 : |
| | : | | | : |
+-----+-----+-----+ +-----+-----+-----+
|4 : : | |4 : : |
| : : | | : : |
+-----+-----+-----+ +-----+-----+-----+
Across:
1. The sum of the digits of 1D. (2)
The product of the digits of 1D. (2)
3. A multiple of 3A from the other figure. (2)
A divisor of 3A from the other figure. (2)
4. A multiple of 1D but not equal to 1D. (3)
Not a multiple of 1D. (3)
Down:
1. Equal to a prime x 3D. (3)
Not equal to a prime x 3D. (3)
2. Differs from the square of 3A by 1. (3)
Differs from the square of 3D by 1. (3)
3. The sum of the digits of 4A. (2)
The product of the digits of 4A. (2)
Please send in the two completed grids
(in either order).
Ciao,
Chappy.
Let 1Aa refer to the first 1a clue, 1Ab refer to the second, and similarly.
3Da must have initial digit 1.
Only ways to have 2Da, 3A ( 3A^2 +- 1 has middle digit match last of 3A)
401 / 20
530 / 23
783 / 28
785 / 28
901 / 30
None of these has an initial digit of 1, so 2Da must correspond to 3Db.
2Db corresponds to 3Da, so has 3Da has initial digit 1:
Only ways to choose 3D st 3D has initial digit 1 and 2Db == 3D^2 +- 1 gives a
value for 3A which is factor or
multiple of (20, 23, 28 or 30):
12 / 143
12 / 145
16 / 255
16 / 257
Using the clue for 3Db gives
1 1 2 2
14 14 15 15
525 723 367 565
In each case, 3D is even and 1D is odd, so 2Db / 3Da corresponds to 1Db,
and 2Da / 3Db corresponds to 1Da.
Using 3Db with the values from 2Da gives
4 4 7 7 7 7 9 9
20 20 28 32 28 28 30 30
641 551 243 713 155 405 661 751
In none of these cases can 4A be a multiple of 1D. (In most, 1d is a
multiple of 2 or 5 where 4a is not; in the fourth, 4A would have to be
9*1D, for last digits to work, but isn't a multiple of 9; in the fifth,
it would make 1D < 100.)
That means 2Db / 3Da / 1Db corr to 4Ab, and 2Da / 3Db / 1Da corr to 4Aa.
1 1 2 2
14 14 15 15
525 723 367 565
Only the first case works with the clue for 4Aa, which makes 1D 105 or 175.
105 doesn't work with either clue for 1A, and 175 only works with 1Aa.
131
714
525
That means the other set must have 3A=28 and 1Ab.
7 7 7
28 28 28
243 155 405
For the second and third cases, 1Da is impossible (3D divisible by 5
but 1D not)
For the first case, the multiplier must end in 3 or 8 for 1D to end
in 2, and no prime ends in 8.
24*3 == 72 (too small)
23*13 == 312
24*23 == 552
24*43 == 1032 (too big)
3 7 5 7
128 528
243 243
In the first case, 1A does not work; in the second case it does.
507
528
243
Final solution
507 131
528 714
243 525
.
- References:
- Enigma 1446 - Double entendre
- From: Chappy
- Enigma 1446 - Double entendre
- Prev by Date: Re: A word puzzle
- Next by Date: Re: OT: US vs UK words
- Previous by thread: Enigma 1446 - Double entendre
- Index(es):
Relevant Pages
|
