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CEO
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How many odd three-digit integers greater than 700 are there [#permalink]
 14 Dec 2007, 08:55
How many odd three-digit integers greater than 700 are there such that all their digits are different?
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CEO
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i made up this question from another question.
please correct me if i am wrong.
for digits starting with 7
controlling for the hundreds digit: 1 possibility
controlling for the ones digit: 4 possibilities (5 odd minus the 7 in the hundreds digit)
controlling for the tens digit: 8 possibilities (cant be hundreds or tens digit)
1*8*4 = 32
for digits starting with 8
controlling for the hundreds digit: 1 possibility
controlling for the ones digit: 5 possibilities (5 odd intgers)
controlling for the tens digit: 8 possibilities (cant be hundreds or tens digit)
1*8*5 = 40
for digits starting with 9
controlling for the hundreds digit: 1 possibility
controlling for the ones digit: 4 possibilities (5 odd minus the 9 in the hundreds digit)
controlling for the tens digit: 8 possibilities (cant be hundreds or tens digit)
1*8*4 = 32
32+40+32 = 104
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CEO
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How many EVEN three-digit integers greater than 700 are there such that all their digits are different?
700-799
1*8*4
units digits cannot have digit 0 because it says greater than 700
800-899
1*8*4
units digit add back the zero but cannot have the digit 8
900-899
1*8*5
32+32+40 = 104
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Director
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that looks right to me 
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Intern
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bmwhype2 wrote: How many EVEN three-digit integers greater than 700 are there such that all their digits are different?
700-799
1*8*4
units digits cannot have digit 0 because it says greater than 700
800-899
1*8*4
units digit add back the zero but cannot have the digit 8
900-899
1*8*5
32+32+40 = 104
700-799
7x0 has 8 (710, 720, 730, 740, 750, 760, 780, 790)
7x2 has 8, Same for 7x4, 7x6, 7x8
so for 7 there are 8 + 8*4 = 40
800-899
8x0 has 8 (810, 820, 830, 840, 850, 860, 870, 890)
Same for 8x2, 8x4, 8x6
so for number starting w/ 8 there are 8*4 = 32
900-999
9x0 has 8 (910, 920, 930, 940, 950, 960, 970, 980)
same for 9x2, 9x4, 9x6, 9x8
so for number starting w/9 there are 8*5 = 40
So total = 40 + 32 + 40 = 112
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Intern
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bmwhype2 wrote: controlling for the ones digit: 4 possibilities (5 odd minus the 7 in the hundreds digit)
Why are you controlling for the odds AND the 7?
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Director
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EarlCat wrote: bmwhype2 wrote: controlling for the ones digit: 4 possibilities (5 odd minus the 7 in the hundreds digit) Why are you controlling for the odds AND the 7? because the hundreds digit is 7 and he can't repeat. for 700-799 it's like this:
701, 703, 705, 709
713, 715, 719
721, 723, 725, 729
731, 735, 739
741, 743, 745, 749
751, 753, 759
761, 763, 765, 769
781, 783, 785, 789
791, 793, 795
32 possibilities (same will hold true for 900)
now for 800-899
801, 803, 805, 807, 809
813, 815, 817, 819
821, 823, 825, 827, 829
831, 835, 837, 839
841, 843, 845, 847, 849
851, 853, 857, 859
861, 863, 865, 867, 869
871, 873, 875, 879
891, 893, 895, 897
40 possibilities
32+40+32 = 104
BMW is correct without going through the tedious chore of writing them all out
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CEO
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ho_ty wrote: bmwhype2 wrote: How many EVEN three-digit integers greater than 700 are there such that all their digits are different?
700-799
1*8*4
units digits cannot have digit 0 because it says greater than 700
800-899
1*8*4
units digit add back the zero but cannot have the digit 8
900-899
1*8*5
32+32+40 = 104 700-799 7x0 has 8 (710, 720, 730, 740, 750, 760, 780, 790) 7x2 has 8, Same for 7x4, 7x6, 7x8 so for 7 there are 8 + 8*4 = 40 800-899 8x0 has 8 (810, 820, 830, 840, 850, 860, 870, 890) Same for 8x2, 8x4, 8x6 so for number starting w/ 8 there are 8*4 = 32 900-999 9x0 has 8 (910, 920, 930, 940, 950, 960, 970, 980) same for 9x2, 9x4, 9x6, 9x8 so for number starting w/9 there are 8*5 = 40 So total = 40 + 32 + 40 = 112 thanks for correcting me.
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