How many 3 digit numbers are possible if the leftmost digit cannot be zero
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How many three-digit odd numbers are possible if the leftmost digit cannot be zero? Two ways: First way: There are 9 choices for the first digit: 1,2,3,4,5,6,7,8,9 There are 10 choices for the second digit: 0,1,2,3,4,5,6,7,8,9 There are 5 choices for the third digit: 1,3,5,7,9 That's 9�10�5 = 450 --------------------------------------- Second way: That's the arithmetic sequence 101,103,...,997,999 with a1 = 101, an = 999, and d=2 an = a1 + (n-1)d 999 = 101 + (n-1)2 Solve for n, get n = 450. Edwin How many digit even numbers are possible if the leftmost digit Cannot be zero?Therefore we can conclude that there are 45,000 possible five digit even numbers are possible. If the left most digit cannot be zero.
How many 3 digit pin can you make if the first digit Cannot be zero?Otherwise, the answer is 18. There are 9 choices for the 1st digit, because it can't be 0. But each of the next 2 choices has 10 options 1–9 + 0.
How many 6 digit odd numbers are possible if the leftmost digit Cannot be 0?5 options for the final (sixth) digit. Then 8 options for the first digit (since it can't be 0, and one of the odds is taken for the final). Then 8 options for the second digit, 7 for the third, 6 for the fourth, 5 for the fifth. 5*8*8*7*6*5=67,200.
How many odd three digit numbers are there assuming that leading 0's are permitted?Assuming leading zeroes are not allowed, eg, 025, (tho it should be specified, we should not have to make assumptions), the 100's digit for 3 digit numbers must be 1,2,3,4,or 5, but one of the odd numbers is used, so it's 1 of 4 for the 100's digit. That's 3*4 = 12 possible 3 digit odd numbers.
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