Can anyone help me solve this puzzle? I could not establish a co-relation between the numbers:
13 = 1001
19 = 10013
23 = 100004
A = 1000001
139 = B
Find the values of A and B.
8
$\begingroup$
$\endgroup$
-
5$\begingroup$ A is 1000001, and B is 139 BOOM! :P $\endgroup$ – Beastly Gerbil May 17 '17 at 6:25
-
$\begingroup$ Has a correct answer been given? If so, please don't forget to $\color{green}{\checkmark \small\text{Accept}}$ it :) $\endgroup$ – Rubio♦ May 19 '17 at 5:24
5
$\begingroup$
$\endgroup$
The left hand side are various prime numbers in increasing order, and they divide the right hand side. Also, the right hand side increases in length by one digit each time. So I think the rule is:
The rhs of the n-th line is the smallest number with n+3 digits that is divisible by the prime on the lhs.From the factorization $1000001 = 101 × 9901$ we get that $A = 101$.
From $1000000 \equiv 62 \mod 139$ we get that $B = 1000000-62+139 = 10000077$.The answer for $A$ could be $9901$ too, but then the numbers wouldn't be strictly increasing.
answered May 17 '17 at 7:05
-
$\begingroup$ That was the correct answer! Thank you!! $\endgroup$ – Amitabh Ghosh May 19 '17 at 16:12
5
$\begingroup$
$\endgroup$
Lets see what we can get here -
If we divide all the prime numbers in group of 10, we get something like this -
2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71 73 79 83 89 97 101 103 107 109 113 127 131 137 139 149 151 157 163 167 173 179 181 191 193 197 199 211 223 227 229 233 239 241 251 257 263 269 271 277 281 283 293 307 311 313 317 331 337 347 349 353 359 367 373 379 383 389 397 401 409 419 421 431 433 439 443 449 457 461 463 467 479 487 491 499 503 509 521 523 541 547 557 563 569 571 577 587 593 599 601 607 613 617 619 631 641 643 647 653 659 661 673 677 683 691 701 709 719 727 733 739 743 751 757 761 769 773 787 797 809 811 821 823 827 829 839 853 857 859 863 877 881 883 887 907 911 919 929 937 941 947 953 967 971 977 983 991 997 1009 1013 1019 1021 1031 1033 1039 1049 1051 1061 1063 1069 1087 1091 1093 1097 1103 1109 1117 1123 1129 1151
Now, if look at the prime factors of each number, we get -
$1001 = 7 × 11 × 13$
$10013 = 17 × 19 × 31$
$100004 = 2^2 × 23 × 1087$
The answer here is the highest prime factor which is not present in nth column where n is 1 or multiple of 11 i.e. it should not be present in the first column of the above box. So, going by that rule we get -
$1001 = 7 × 11 × 13$ $\rightarrow$ 13
$10013 = 17 × 19 × 31$ $\rightarrow$ 19
$100004 = 2^2 × 23 × 1087$ $\rightarrow$ 23
Hence, value of A can be deduced as -
$1000001 = 101 × 9901$ $\rightarrow$ 101 (Since 9901 is in the first column and can be dropped) [Reference]
Hence, A = 101
So, if we go by this rule, we can have multiple values for B.
-
$\begingroup$ Well that's.... complicated. And for B it's the other way round, so 139 is the highest prime factor not in the nth column? $\endgroup$ – Beastly Gerbil May 17 '17 at 6:47
-
$\begingroup$ @BeastlyGerbil For B, 139 is the LHS answer. So, we need to look for numbers whose factors include 139 and another higher prime number than 139 which rests in 1st column. So, it could be anything like $1391251 = 139*10009$ making B = 1391251. And that's why I said, B can have many values if we follow this rule. Also, I think, that was the simplest rule I could have came up with(considering how bad I am at math) ;) $\endgroup$ – Techidiot May 17 '17 at 6:50
