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How many 5-digit numbers can be constructed that both begin and end with the digit 3 and contain unlimited repetitions?
Wiki User
∙ 8y ago
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Lenora Corkery
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∙ 8y agoAs the first and last digits are fixed to be 3, only the middle 3 digits can change. As repetitions are allowed, each of these digits can be any of the 10 digits {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}. There is a choice of 10 digits for the first, and for each of these choices there is a choice of 10 for the second and for each of these choices there is a choice of 10 for the last digit; thus there are 10 × 10 × 10 = 1000 such possible numbers.
Wiki User
∙ 8y ago1 option for the first digit, 1 option for the last digit, 10 options for the second digit, 10 options for the third digit, 10 options for the fourth digit. Multiply all these numbers together.
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Sample code of loops and repetitions?
system.out.println(" print 1-100 numbers");for(i=0;i
