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 Post subject: Re: Three Digit Integers
PostPosted: Thu Nov 19, 2009 8:43 am 
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Joined: Mon Apr 06, 2009 5:44 pm
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If I recall correctly, this is a problem from the official guide. It isn't as bad as it looks. Our set of numbers is 701-999, we are looking for numbers of the form xxy, yxx, and xyx two of the digits, the x's are identical, and the other digit, y, is different. Lets make some lists and begin with our 7's

with the 7's we have 707, 717, ..., 797 that's 10 but we can't have 777 since all the digits are the same, so we actually have 9 numbers of the form xyx where x =7. Next we have the numbers where we have yxx. So, we have 711, 722, 733, ..., 799 again we have 9 but we have to remove 777, so there are actually 8. Next we look at the numbers of the form xxy. So we have 771, 772, .....779. again we remove the 777 and we have 9. So there are 9+ 9 + 8 = 26 such numbers in the 7's

with the 8's we have 808, 818, ..., 898 that's 10 but we can't have 888 since all the digits are the same, so we actually have 9 numbers of the form xyx where x =8. Next we have the numbers where we have yxx. So, we have 800, 811, 822, 833, ..., 899 again we have 10 but we have to remove 888, so there are actually 9. Next we look at the numbers of the form xxy. So we have 881, 882, .....889. again we remove the 888 and we have 9. So there are 9 x 3 = 27 such numbers in the 8's

with the 9's we have 909, 919, ..., 999 that's 10 but we can't have 999 since all the digits are the same, so we actually have 9 numbers of the form xyx where x =9. Next we have the numbers where we have yxx. So, we have 900, 922, 933, ..., 999 again we have 10 but we have to remove 999, so there are actually 9. Next we look at the numbers of the form xxy. So we have 991, 992, .....999. again we remove the 999 and we have 9. So there are 9 x 3 = 27 such numbers in the 9's

so we have 27 + 27 + 26 = 80


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