The Eukaryotic Professor Has A M¨obius Bicycleloc

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The eukaryotic Professor has a M¨obius bicyclelock with a tendigit combination. To remember the combination, the Professormade up a codeword and a list of clues. The codewordis HYPERBOLIC and each letter represents a different digit. The cluesare (1) O + O +O = C (2) P / B =Y (3) E / O =Y (4) B + B =P (5) O × B= RY (6) O / E =B / P (7) L + P +I = C + H +R (8) H × B= YI (9) E + H =RR (10) C × O= YL (RY, YI,RR, and YL are two-digitnumbers). What isthe ten-digit combination?

 

Answer No.1

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Begin with #2 and #4. #2 states that P/B=Y. #4 states that B+B=P. So 2B/B=Y, and from that you get Y=2. Then take a took at #1 and #10. #1 says O+O+O=C. #10 saysC*O=YL. So then 3O*O=YL. And Y = 2, so 3O*O= 20 + L. To isolate the O's,divide by 3. Now you have O^2=(20+L)/3. From this you can logically concludethat L is 7(If you wanted to test all your choices, you would onlyneed to bother with 1, 4, and 7 anyway). Test it to find O. O^2=27/3. 9=O^2, O=3. Look back at the first one. O+O+O=C. 3+3+3= C. C is 9. Now, find E using #3. E/O=Y. E/3=2, so E=6. #9: E+H=RR. 6+H=RR. This hard, and you'll hind out that you add 5.6+5=11, thus, H=5 and R=1. #5 says O*B=RY. 3*B=12, and B comes out to be 4. Both #'s 4 and 2 will tell you P. 4+4=P, P=8. And if you use #2,P/4=2, and again, P=8. #10 says that C*O=YL. 9*3=20+L. 27=20+L, so L=7. Now, the only one left is I. You could use process of eliminationor use #8. H*B=YI. 5*4=20+I. I=0. H=5, Y=2, P=8, E=6, R=1, B=4, O=3, L=7, I=0, C=9. 5286143709.

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