Posted 6 Jun 2014 At the aforementioned k-man took only imperceptible offense at Y-san's mainly-innocent comment. But he did retire brieflly to the library to concoct some good-natured revenge on the lady. Returning moments later to the main room he found her in conversation with Rainman, and he put to them a puzzle of his own: I've put a number into a sealed envelope here, he said, and it looks like this. And he showed them the sequence ABCDCBA. But of course my copy has numbers for the letters, he explained. I'm going to tell Y-san the sum of the seven digits, and I'll tell Rainman their product. Bonanova is not the only one who can resurrect an old puzzle form - you may already have seen puzzles of this type. It will be helpful if you have. You can discuss what you know between yourselves, but be careful not to disclose too much. First to come up with my seven-digit number gets a coveted bonanova gold star. Then k-man added, here's a clue: 9 >= A >= B >= C >= 0. and 9 >= D >= 0. Y-san spoke first, to Rainman: I have only a faint clue about what your number is. I have an even worse idea of what yours is, Rainman replied, this is difficult. I know, Y-san mused. At this point, your number is still uncertain. Rainman replied, now I know k-man's number! What is it? 0 Share this post Link to post Share on other sites

0 Posted 9 Jun 2014 Nice work. But since Rainman knows the product, he knows at least one of these is not the right number. 0 Share this post Link to post Share on other sites

0 Posted 9 Jun 2014 Nice work. But since Rainman knows the product, he knows at least one of these is not the right number. Since, Rainman knows the product, he doesn't even have to think about the other number. This doesn't change the fact that there are indeed two answers to your problem. Or, am I wrong in stating that there are two possible answers to your question? 0 Share this post Link to post Share on other sites

0 Posted 9 Jun 2014 I think there is only one answer. Here are the details of what I did. See if I missed something. You said, since Rainman knows the product, he doesn't even have to think about the other number. The problem is that, if there are two possible numbers, he doesn't know the answer. Consider the two suggested answers: 1114111 and 9869689. They have sums of 10 and 55 respectively They have products of 4 and 1679616 respectively. InitiallyThe 2200 possible ABCDCBA numbers have 64 different sums, from 0 to 63.9 of them, 0 1 2 3 4 5 6 62 and 63 correspond to unique products. They are, respectively, 0 0 0 0 0 0 0 4251528 and 4782969. Since Y-san can't determine the product, none of these are her sum. We can eliminate numbers having those sums, leaving 2177 possibilities. After Y-san's first statement The 2177 possible numbers have 679 different products.327 of them, that do not include 4 or 1679616, correspond to unique sums. Since Rainman can't determine the sum, none of them are his product. We can eliminate numbers having those "unique" products, leaving 1844 possible numbers. Since 4 and 1679616 were not eliminated, 1114111 and 9869689 are still possible. After Rainman's first statement The 1844 possible numbers have 52 different sums.5 of them, 7 8 56 57 58, correspond to unique products. The products are, respectively, 0 0 1679161 2125764 and 2125764. Since Y-san still can't determine the product, none of these are her sum. We can eliminate numbers having those sums, leaving 1825 possible numbers. After Y-san's second statement The 1825 possible numbers have 351 different products At this point, only 1 of them, 1679616, corresponds to a distinct sum, 55.Rainman says I know k-man's number So Rainman must have been given the product of 1679616. So (product=1679616) points to only one number9869689 sum=55. While (product=10) points to two different numbers that have different sums.1114111 sum=102111112 sum=9 ========== Looking closer, Before Y-san's second statement, (product=1679616) pointed to two different numbers.(1679616=TABLES1[;7])/[1]TABLES1 <- this asks for rows where col7 (prod)=1679616 9 8 6 9 9869689 55 1679616 9 9 8 4 9984899 56 1679616 <- so the number 9984899 was possible at this point. Then Y-san's second statement eliminated (sum=56), (see red above) making 9869689 unique.Also, before Y-san's second statement, (product=4) pointed to two different numbers.(4=TABLES2[;7])/[1]TABLES2 <- this asks for rows where col7 (prod)=4 1 1 1 4 1114111 10 4 2 1 1 1 2111112 9 4 Y-san's second statement did not eliminate (sum=9) or (sum=10). This leaves (product=4) ambiguous. So Rainman's product could not be 4. In any event, I'm marking the puzzle solved. Good work. 0 Share this post Link to post Share on other sites

0 Posted 10 Jun 2014 I think there is only one answer. Here are the details of what I did. See if I missed something. You said, since Rainman knows the product, he doesn't even have to think about the other number. The problem is that, if there are two possible numbers, he doesn't know the answer. Consider the two suggested answers: 1114111 and 9869689. They have sums of 10 and 55 respectively They have products of 4 and 1679616 respectively. InitiallyThe 2200 possible ABCDCBA numbers have 64 different sums, from 0 to 63.9 of them, 0 1 2 3 4 5 6 62 and 63 correspond to unique products. They are, respectively, 0 0 0 0 0 0 0 4251528 and 4782969. Since Y-san can't determine the product, none of these are her sum. We can eliminate numbers having those sums, leaving 2177 possibilities. After Y-san's first statement The 2177 possible numbers have 679 different products.327 of them, that do not include 4 or 1679616, correspond to unique sums. Since Rainman can't determine the sum, none of them are his product. We can eliminate numbers having those "unique" products, leaving 1844 possible numbers. Since 4 and 1679616 were not eliminated, 1114111 and 9869689 are still possible. After Rainman's first statement The 1844 possible numbers have 52 different sums.5 of them, 7 8 56 57 58, correspond to unique products. The products are, respectively, 0 0 1679161 2125764 and 2125764. Since Y-san still can't determine the product, none of these are her sum. We can eliminate numbers having those sums, leaving 1825 possible numbers. After Y-san's second statement The 1825 possible numbers have 351 different products At this point, only 1 of them, 1679616, corresponds to a distinct sum, 55. Rainman says I know k-man's number So Rainman must have been given the product of 1679616. So (product=1679616) points to only one number9869689 sum=55. While (product=10) points to two different numbers that have different sums.1114111 sum=102111112 sum=9 ========== Looking closer, Before Y-san's second statement, (product=1679616) pointed to two different numbers.(1679616=TABLES1[;7])/[1]TABLES1 <- this asks for rows where col7 (prod)=1679616 9 8 6 9 9869689 55 1679616 9 9 8 4 9984899 56 1679616 <- so the number 9984899 was possible at this point. Then Y-san's second statement eliminated (sum=56), (see red above) making 9869689 unique.Also, before Y-san's second statement, (product=4) pointed to two different numbers.(4=TABLES2[;7])/[1]TABLES2 <- this asks for rows where col7 (prod)=4 1 1 1 4 1114111 10 4 2 1 1 1 2111112 9 4 Y-san's second statement did not eliminate (sum=9) or (sum=10). This leaves (product=4) ambiguous. So Rainman's product could not be 4. In any event, I'm marking the puzzle solved. Good work. First of all, thanks for the star. Its been decades since I got one Now,.."The 1844 possible numbers have 52 different sums. 5 of them, 7 8 56 57 58, correspond to unique products." Actually, 7 and 8 are eliminated earlier as both correspond to unique numbers 1111111 and 1112111. So rainman would have guessed them straight away as he would have got the product 1 or 2, respectively. Your statement should read: "4 of them, 9 56 57 58, correspond to unique products." The sum 9 gets eliminated here. 9 gives two products-3 and 4. If rainman has 3, he will tell the sum, immediately after y-san's first statement. ALTERNATIVELY - you state "327 of them, that do not include 4 or 1679616, correspond to unique sums". 3 happens to get eliminated here as it has a unique sum 9. Therefore after Rainman's 1st statement the sum 9 corresponds to only 1 unique product 4. It has to be eliminated otherwise Y-San will know rainman's product (and the number). Thus "4 of them, 9 56 57 58, correspond to unique products." 0 Share this post Link to post Share on other sites

0 Posted 10 Jun 2014 There are four points in time, separated by the first three statements.Initially Nine sums, 0 1 2 3 4 5 6 62 and 63 point to unique products, keeping Y-san from knowing the product, and these sums are eliminated. Note that (sum=7) and (sum=8) are not among the eliminated possibilities, since the groups of numbers they point to associate with multiple products (0,1) and (0,2) respectively. Initially (sum=7) refers to seven numbers:(7=TABLE[;6])/[1] TABLE 0 0 0 7 7000 7 0 1 0 0 5 1005001 7 0 1 1 0 3 1103011 7 01 1 1 1 1111111 7 1 2 0 0 3 2003002 7 0 2 1 0 1 2101012 7 0 3 0 0 1 3001003 7 0Initially (sum=8) refers to eleven numbers(8=TABLE[;6])/[1] TABLE 0 0 0 8 8000 8 0 1 0 0 6 1006001 8 0 1 1 0 4 1104011 8 01 1 1 2 1112111 8 2 2 0 0 4 2004002 8 0 2 1 0 2 2102012 8 0 2 1 1 0 2110112 8 0 2 2 0 0 2200022 8 0 3 0 0 2 3002003 8 0 3 1 0 0 3100013 8 0 4 0 0 0 4000004 8 0 They would be eliminated only in case all the numbers they point to have a single value for their associated products. At this point that is not the case.After Y-san's first statement Y-san's first statement removes 327 products that refer to (groups of) numbers for which there is a single sum. They are: 1 2 3 5 6 7 12 18 20 24 25 27 28 45 49 50 54 63 75 98 125 147 150 175 245 294 300 343 450 500 588 600 625 675 700 882 980 1125 1176 1225 1250 1323 1350 1372 1575 1875 2205 2401 2450 2646 3087 3125 3675 3750 4375 4802 6125 7203 7350 7500 8575 11250 12005 12500 14406 14700 15000 15625 16807 16875 17500 22050 24500 28125 28812 29400 30000 30625 31250 33075 33750 34300 36450 39375 43218 45000 46875 48020 50000 54675 55125 57624 58800 60000 60025 61250 62500 64827 66150 67228 67500 70000 71442 76800 77175 78125 88200 91125 91875 93750 98000 101250 107163 108045 109350 109375 112500 115248 117600 117649 118098 120000 120050 122500 125000 127575 128000 129654 132300 135000 137200 140625 145800 150528 151263 151875 153125 153600 157500 160000 172800 172872 177147 178605 179200 180000 180075 183750 192080 196608 198450 200000 214326 214375 218700 220500 230496 235200 235298 240000 240100 245000 250047 250880 253125 259308 264600 268912 275625 280000 285768 288000 295245 297675 300125 301056 303750 307200 308700 313600 320000 327680 328050 338688 345600 351232 352800 352947 354294 354375 360000 360150 364500 388800 388962 392000 393216 403200 405000 409600 413343 420175 428652 432180 437400 442368 455625 458752 460800 460992 470400 470596 472392 480200 492075 496125 510300 512000 518616 540225 548800 559872 564480 583200 583443 588245 595350 602112 605052 614400 614656 627200 642978 648000 677376 691488 694575 705600 705894 708588 714420 716800 737280 762048 768320 777600 786432 790272 793800 802816 819200 820125 823543 857304 884736 893025 903168 907200 921600 921984 933120 941192 964467 972405 984150 995328 1000188 1003520 1032192 1036800 1048576 1058841 1062882 1075648 1119744 1143072 1148175 1166400 1166886 1179648 1180980 1204224 1229312 1259712 1270080 1306368 1310720 1312200 1361367 1382976 1404928 1417176 1476225 1492992 1524096 1555848 1572864 1594323 1605632 1607445 1653372 1658880 1750329 1778112 1806336 1835008 1889568 1928934 1990656 2032128 2097152 2099520 2250423 2286144 2322432 2359296 2519424 2571912 2654208 2657205 2893401 2939328 2985984 3188646 3359232 3720087 3779136 Specifically, 1 and 2 are among the products eliminated at this point. (prod=1) and (prod=2) refer to these numbers:(1=TABLES1[;7])/[1] TABLES1 1 1 1 1 1111111 7 1 (2=TABLES1[;7])/[1] TABLES1 1 1 1 2 1112111 8 2 As you point out, (product=3) is also among them. It points to a single number, 1113111, which gives a unique sum of 9. (I wonder whether you are eliminating (sum=9) from the table here? Probably not, but you mention (sum=9) later.(3=TABLES1[;7])/[1] TABLES1 1 1 1 3 1113111 9 3 But (product=4) and (product=1679616) are not eliminated here: their associated sums are not unique.(4=TABLES1[;7])/[1] TABLES1 Both 1114111 and 2111112 are still possible1 1 1 4 1114111 10 4 2 1 1 1 2111112 9 4(1679616=TABLES1[;7])/[1] TABLES1 Both 9869689 and 9984899 are still possible9 8 6 9 9869689 55 1679616 9 9 8 4 9984899 56 1679616 After Rainman's first statement There are 52 different sums, and exactly 5 of them, 7 8 56 57 58 have unique products:(7=TABLEP1[;6])/[1] TABLEP1 (sum=7) is eliminated now, not before.0 0 0 7 7000 7 0 1 0 0 5 1005001 7 0 1 1 0 3 1103011 7 0 2 0 0 3 2003002 7 0 2 1 0 1 2101012 7 0 3 0 0 1 3001003 7 0 (8=TABLEP1[;6])/[1] TABLEP1 (sum=8) is eliminated now, not before.0 0 0 8 8000 8 0 1 0 0 6 1006001 8 0 1 1 0 4 1104011 8 0 2 0 0 4 2004002 8 0 2 1 0 2 2102012 8 0 2 1 1 0 2110112 8 0 2 2 0 0 2200022 8 0 3 0 0 2 3002003 8 0 3 1 0 0 3100013 8 0 4 0 0 0 4000004 8 0 (56=TABLEP1[;6])/[1] TABLEP1 9 9 8 4 9984899 56 1679616 (57=TABLEP1[;6])/[1] TABLEP1 9 9 6 9 9969699 57 2125764 (58=TABLEP1[;6])/[1] TABLEP1 9 9 9 4 9994999 58 2125764 You state that (sum=9) should be eliminated at this point. But at this point, (sum=9) does not qualify for elimination - its associated products are not unique.:(9=TABLEP1[;6])/[1] TABLEP1 0 0 0 9 9000 9 0 1 0 0 7 1007001 9 0 1 1 0 5 1105011 9 0 2 0 0 5 2005002 9 0 2 1 0 3 2103012 9 02 1 1 1 2111112 9 4 2 2 0 1 2201022 9 0 3 0 0 3 3003003 9 0 3 1 0 1 3101013 9 0 4 0 0 1 4001004 9 0 You state that "after Rainman's 1st statement the sum 9 corresponds to only 1 unique product 4. It has to be eliminated otherwise Y-san will know rainman's product (and the number). Thus "4 of them, 9 56 57 58, correspond to unique products."It seems that this is the point of departure for our analyses. If you take the number 2111112 out of play here, you leave 1114111 as a unique number for (product=4) I think you can't take 2111112 out of play at this point because (sum=9) does not correspond to a unique product.After Y-san's second statement I find that among the 351 different remaining products, [length of] 2 REMDUP [remove duplicates of] TABLES2[;7]351 only one product refers to a unique sum:REMUPRODS TABLES2 351 PRODS occur in the TABLE. 1 of them point to unique SUMS. UPRODS = 1679616 And that number is (1679616=TABLES2[;7])/[1]TABLES2 9 8 6 9 9869689 55 1679616 Just occurred to me. Did you permit zeros? 0 Share this post Link to post Share on other sites

0 Posted 14 Jun 2014 Yes Bonanova, I did not permit zeros as I thought in that case Rainman would have given a different initial reply (that he can never find out Y-San's product) 0 Share this post Link to post Share on other sites

0 Posted 14 Jun 2014 Ah that explains it. Good solve. 0 Share this post Link to post Share on other sites

Posted

At the aforementioned k-man took only imperceptible offense at

Y-san's mainly-innocent comment. But he did retire brieflly to the library to concoct

some good-natured revenge on the lady. Returning moments later to the main room

he found her in conversation with Rainman, and he put to them a puzzle of his own:

I've put a number into a sealed envelope here, he said, and it looks like this.

And he showed them the sequence

. But of course my copy has numbersABCDCBAfor the letters, he explained.

I'm going to tell Y-san the sum of the seven digits, and I'll tell Rainman their product.

Bonanova is not the only one who can resurrect an old puzzle form - you may already

have seen puzzles of this type. It will be helpful if you have. You can discuss

what you know between yourselves, but be careful not to disclose too much.

First to come up with my seven-digit number gets a coveted bonanova gold star.

Then k-man added, here's a clue: 9 >=

>=A>=B>= 0. and 9 >=C>= 0.DY-sanspoke first, to Rainman: I have only a faint clue about what your number is.I have an even worse idea of what yours is,

Rainmanreplied, this is difficult.I know,

Y-sanmused. At this point, your number is still uncertain.Rainmanreplied, now I know k-man's number!What is it?

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