Rob_Gandy 2 Posted January 10, 2013 Report Share Posted January 10, 2013 How many 3-digit positive integers are odd and do not contain the digit 5? I got 320, but the answer that is given in 288. I just realized that I missed some numbers, but I ended up with 280. I would greatly appreciate if someone could explain why the answer is 288. Quote Link to post Share on other sites

Solution Thalia 64 Posted January 10, 2013 Solution Report Share Posted January 10, 2013 (edited) There are 900 3-digit numbers (100-999) Half of them are odd: 900/2=450 There are 9 groups of hundreds (100s, 200s, etc) Ignoring the 500s for the moment, each group of 100 has 14 odd numbers with a 5 (5, 15, 25, 35, 45, 51, 53, 55, 57, 59, 65, 75, 85, 95) 8 (groups of 100 ignoring the 500s) * 14=112 450-112=338 All of the numbers in the 500s have a 5. Half of them are odd: 100/2=50 338-50=288 Edited January 10, 2013 by Thalia Quote Link to post Share on other sites

Rob_Gandy 2 Posted January 10, 2013 Author Report Share Posted January 10, 2013 Thank you much. I must have done some miscounting along the way. Quote Link to post Share on other sites

Kay_kay 1 Posted May 7, 2015 Report Share Posted May 7, 2015 Its a three digit no., so let's say it is ABC (C - units place, B - tens place, A - hundreds place) possible values for each: C: 1, 3, 5, 7, 9 [4 possible values] B: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 [9 possible values] A: 1, 2, 3, 4, 5, 6, 7, 8, 9 [8 possible values] So, possible number combinations to give required 3 digit nos. :4 x 9 x 8 = 288 1 Quote Link to post Share on other sites

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