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(a) the decimal equivalent of (12LM.L1);

(b) the total number of possible four-digit combinations in this arbitrary number system.

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Solution

(a) The decimal equivalent of (12LM) is given by

M×50 +L×51 +2×52 +1×53 = 4×50 +3×51 +2×52 +1×53

L = 3M = 4

= 4+15+50+125 = 194

The decimal equivalent of (L1) is given by

L×5−1 +1×5−2 = 3×5−1 +5−2 = 0.64 Combining the results, (12LM.L1)5 = (194.64)10.

(b) The total number of possible four-digit combinations = 5*5*5*5 = 625.

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