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How many odd three-digit integers greater than 700 are there

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CEO
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How many odd three-digit integers greater than 700 are there [#permalink] New post 14 Dec 2007, 07:55
How many odd three-digit integers greater than 700 are there such that all their digits are different?
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 [#permalink] New post 14 Dec 2007, 08:17
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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 [#permalink] New post 14 Dec 2007, 08:37
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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 [#permalink] New post 14 Dec 2007, 08:37
that looks right to me 8-)
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 [#permalink] New post 14 Dec 2007, 19:10
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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 [#permalink] New post 14 Dec 2007, 23:08
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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 [#permalink] New post 15 Dec 2007, 08:55
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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 [#permalink] New post 16 Dec 2007, 18:19
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.
  [#permalink] 16 Dec 2007, 18:19
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