Geodood 0 Report post Posted August 23, 2018 Hi everyone! Today I came across with this riddle that seems simple, but I haven't got a clue on how to solve. IF 3+5+7 = 152131 4+8+6 = 322448 6+2+9 = 125464 THEN A + B + C = 123040 -------------------------- IF 9 + 4 + 5 = 364590 2 + 6 + 8 = 121630 3 + 7 + 8 = 212448 THEN D + E + F = 303570 Thanks in advance Quote Share this post Link to post Share on other sites
0 Thalia 64 Report post Posted August 24, 2018 A start: Given A+B+C, the first two digits are the product of AB. The next two digits are the product of AC. Still working on the last two digits. Quote Share this post Link to post Share on other sites
2 Thalia 64 Report post Posted August 24, 2018 Full solution: Given A+B+C, the first two digits are the product of AB. The next two digits are the product of AC. The last two digits are AB+AC-B. For D+E+F, the last two digits are DE+DF+A. A=6 B=2 C=5 D=5 E=6 F=7 Quote Share this post Link to post Share on other sites
0 sushma 0 Report post Posted August 24, 2018 Hi Geodood, logic is : 3+5+7 = 152131 solved as (3*5) (3*7) (15+21-5) In shorts we can say abc = (a*b)(a*C)(a*b+a*c-b) Then A+B+C = 123040 In this A = 6 B=2 C=5 Similarly for IF 9 + 4 + 5 = 364590 2 + 6 + 8 = 121630 3 + 7 + 8 = 212448 here A+B+C = (A*B) (A*C) (A*B+A*C+A) so THEN D + E + F = 303570 D = 5 E = 6 F = 7 Quote Share this post Link to post Share on other sites
0 Geodood 0 Report post Posted August 26, 2018 I say thanks to all of you who gave a little hand! Quote Share this post Link to post Share on other sites
Hi everyone!
Today I came across with this riddle that seems simple, but I haven't got a clue on how to solve.
IF
3+5+7 = 152131
4+8+6 = 322448
6+2+9 = 125464
THEN
A + B + C = 123040
--------------------------
IF
9 + 4 + 5 = 364590
2 + 6 + 8 = 121630
3 + 7 + 8 = 212448
THEN
D + E + F = 303570
Thanks in advance
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