Geodood Posted August 23, 2018 Report Share Posted August 23, 2018 (edited) 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 Edited August 23, 2018 by Geodood Quote Link to comment Share on other sites More sharing options...
0 vrt Posted August 24, 2018 Report Share Posted August 24, 2018 The first 2 digits are the product of the first 2 numbers in the sum: 3*5=15. Then you multiply the first and last numbers: 3*7=21. Finally you add 15+21 and subtract the centre number: 15+21=36, 36-5=31. So if A+B+C=123040, then AB=12, AC=30, and B=42-40=2, so A=6, C=5. In the second version, instead of subtracting the 2nd number, you're adding the first: 3*7=21, 3*8=24, 21+24+3=48. So if D+E+F=303570, DE=30, DF=35, and 65+D=70, making D=5, E=6, F=7. Quote Link to comment Share on other sites More sharing options...
0 Geodood Posted August 26, 2018 Author Report Share Posted August 26, 2018 I say thanks to all of you who gave a little hand! Quote Link to comment Share on other sites More sharing options...
Question
Geodood
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
Edited by GeodoodLink to comment
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