Cryptarithmetic Problem 3
Solution:
T E A x H A D L D T R H R S A E W D A L E S S E R Here, A x A = _ A [ H R S A ] Therefore, Possible values of A= {5, 6} Detailed Explanation- Rule 2 Further, T E A x H A D L D T R H R S A E W D A L E S S E R Here, H x A= _A [ E W D A ] Therefore, two possible cases for the values of H and A Case I - when A={5} then H={3, 7, 9} Case II - when A={2, 4, 8} then H={6} Detailed Explanation-Rule 3 Firstly taking case - I Taking A=5 rewrite the problem again, T E 5 x H 5 D L D T R H R S 5 E W D 5 L E S S E R Further, T E 5 x H 5 D L D T R H R S 5 E W D 5 L E S S E R Here, 5 x D = _ R [ L D T R ] Now, you can easily predict the value of R = 0 and possible values of D= {2, 4, 6, 8} [If you multiply 5 to a number, you will only get[0,5] as their unit digit.] 5 x Even =_0 [2, 4, 6, 8] 5 x Odd =_5 [3, 5, 7, 9] [Relax it's going to take some to understand the concept. Please read... again!] Put R=0 and write the problem again, T E 5 x H 5 D L D T 0 H 0 S 5 E W D 5 L E S S E 0 At this stage, divide the problem into 3 parts, (1) T E 5 x D L D T 0 (2) T E 5 x 5 H 0 S 5 (3) T E 5 x H E W D 5 Now, take (2) [ As it has less number of variables. 5 is repeated three times.] (2) T E 5 x 5 H 0 S 5 Now you have to start hit and trial with the possible values of E Firstly take E=1 Put E=1 in (2) (2) T 1 5 x 5 H 0 S 5 [ H 0 7 5] If you compare side by side, then you will get S=7 Put S=7 and E=1 in the main problem. [It needs to be checked further whether these values satisfies the Basic Cryptarithmetic Rules] T 1 5 x H 5 D L D T 0 H 0 7 5 1 W D 5 L 1 7 7 1 0 At this stage you can easily predict all the values as You can see T + 5 =_1 (which is only possible when the value of the T=6) L=2 (As, 1 + 1(carry) = L) Hence T=6, L=2. Now you can easily solve the problem. 6 1 5 x 3 5 4 2 4 6 0 3 0 7 5 1 8 4 5 2 1 7 7 1 0
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