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ssrs barcode font download x = 3; Module[{x = 8}, x + 1] 9 x 3 in Software
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x=3 3 Module[{x}, Print[x]] x$342 Module[{x}, Print[x]] x$344 Module[{x}, Print[x]] x$346 x 3
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SOLVED PROBLEMS
A.5 Write a module that will take an integer and return all its factors.
SOLUTION
factorlist[x0_] Module[{x = 1}, While[x x0, If[Mod[x0, x] 0, Print[x]]; x++]] factorlist[1] 1 factorlist[10] 1 2 5 10 factorlist[11] 1 11 APPENDIX
factorlist[90] 1 2 3 5 6 9 10 15 18 30 45 90 A.6 A very crude way of determining the position of a prime within the sequence of primes is to examine the list of all primes up to and including the prime in question and determine its position in the list. If the number is not in the list, then the number is not prime. Construct a module that will determine whether a number is prime, and if so, determine its position. If not, return a message indicating that it is not prime. SOLUTION
pos[x0_] Module[{x = 1, prm}, prm = False; While[Prime[x] x0 && Not[prm], If[Prime[x] x0, prm = True]; x ++]; If[prm, Print[x 1], Print["Not a Prime"]]] pos[1] Not a Prime pos[2] 1 pos[3] 2 pos[101] 26 pos[1001] Not a Prime A.7 A famous conjecture asserts that if you start with a positive integer, n, and replace it by n /2 if n is even and by 3 n + 1 if n is odd, and repeat the process over and over in an iterative manner, then you will always wind up with 1. (This conjecture has never been proven or disproved.) Construct a module that simulates this iterative process. SOLUTION
We first define a function, successor, that will define one iteration step. successor[n_] If[EvenQ[n], n/2, 3n + 1] Next we introduce a module, allvalues, that will produce a list of all successors, starting with the successor of n. APPENDIX
allvalues[n_] := Module[{m = n}, While[m 1, m = successor[m]; Print[m]]] allvalues[6] 3 10 5 16 8 4 2 1 Since this list might be long if n is large, and all we are really interested in is the final value and the number of iterations it takes to get there, another module might be more appropriate. finalvalue[n_] Module[{m = n, k = 0}, While[m 1, m = successor[m]; k++]; Print["final value = ", m, ", # iterations = ", k]] finalvalue lists the final value of the process, together with the number of iterations needed to reach the final value. finalvalue[6] final value = 1, # iterations = 8 finalvalue[100] final value = 1, # iterations = 25 finalvalue[1000] final value = 1, # iterations = 111

