You are watching: How many 3-digit positive integers are odd and do not contain the digit 5


Hi rennster:449 is no correct, yet you"re only off through one. Let"s forget about an unfavorable numbers for currently to do the conversation easier; we deserve to always twin our final an outcome later to incorporate the an adverse counterparts.There are various ways to counting the number of 3-digit strange integers; here"s the an approach that popped right into my psychic first.We understand that this collection of integers is 101, 103, 105, ... 999.A formula for all of the odd integers is (displaystyle 2n - 1) as n goes indigenous 1 come infinity.We"re interested in 101, 103, 105, .... We acquire these number by including the odd integers to 100.(displaystyle 100 + 2n - 1) simplifies to (displaystyle 2n + 99)Therefore, a formula for the numbers 101, 103, 105, ... Is(displaystyle 2n + 99) for (displaystyle n = 1, 2, 3, ...)We want our sequence of numbers to stop at 999. What value of n generates the number 999?(displaystyle 2n + 99 = 999)(displaystyle 2n = 900)(displaystyle n = 450)Therefore, there room 450 three-digit weird integers because there is a one-to-one correspondence between the sets 1, 2, 3, ... 450 and 101, 103, 105, ... 999.Perhaps you arrived at 449 instead because you go the following.(displaystyle frac999-1012 = 449)The error in logic with this strategy is the subtraction does not account for the number at each finish of an interval as soon as counting. Subtraction only offers you the difference in between the numbers at each end of an interval.For example, how many integers space there from 1 v 10?You can be tempted to calculate 10 minus 1 and report the there are nine integers indigenous 1 v 10. That"s wrong.
Huh? :shock: it s okay -- possibly you room trying come say something around the reality that after removing every one of the numbers that end in the digit 5, the last digit that the remaining numbers cycle with the other four odd numbers less than 10 (i.e., the critical digit adheres to the pattern 1, 3, 7, 9, 1, 3, 7, 9, 1, 3, 7, 9, ...).If this is what you to be thinking, then you"re ~ above the ideal track.In the formula (displaystyle 2n + 99), the last digit the the resulting numbers cycles through 1, 3, 5, 7, 9 because that every 5 increments of n.n = 1 come 5 ? 101, 103, 105, 107, 109n = 6 to 10 ? 111, 113, 115, 117, 119n = 11 to 15 ? 121, 123, 125, 127, 129n = 16 to 20 ? 131, 133, 135, 137 ,139and for this reason on ...See the pattern? In various other words, over there is exactly one number the ends in the digit 5 created each time n boosts through five values.Therefore, the total variety of times the n boosts through 5 values when going native 1 come 450 is the exact same as the total variety of integers that finish in the number 5 within the set 101, 103, 105, ... 999.It"s very basic to calculate this (i.e., the variety of 5-value boosts that n goes with on its means from 1 come 450).Subtracting the from the number of 3-digit weird integers outcomes in the variety of 3-digit strange integers that carry out not end in the number 5.Doubling this last calculation accounts because that both the negatives and also the positives, and that provides the answer come your original question.So, what carry out you obtain :?: Cheers, ~ Mark

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I Misread Your original Problem
Whoops!Somebody just pointed the end that ns misread your original problem.Somehow, I got the idea in my head that we"re talking around 3-digit integers that finish in 5. Okay, allow me scramble for a couple of minutes to see if I can salvage anything indigenous my an initial post ...(Arrrrgh)