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by goyalsau » Mon Oct 04, 2010 4:47 am
How many odd three-digit integers greater than 800 are there such that all their digits are different?
72
40
56
81
104
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Source: — Problem Solving |

by shovan85 » Mon Oct 04, 2010 5:20 am
IMO A

Hundredth place either 8 or 9
Tenth place 0 to 9
Unit place 1,3,5,7,9

When Hundredth place is 8, Tenth place 0,2,4,6 (Not 8 as repeat), Unit place can be (1,3,5,7,9) = 1*4*5 = 20 Numbers
When Hundredth place is 8, Tenth place 1,3,5,7,9, Unit place can be (1,3,5,7,9 except the current tenth one) = 1*5*4 = 20 Numbers

When Hundredth place is 9, Tenth place 0,2,4,6,8, Unit place can be (1,3,5,7 Not 9) = 1*5*4 = 20 Numbers
When Hundredth place is 8, Tenth place 1,3,5,7(not 9), Unit place can be (1,3,5,7,9 except the current tenth one and 9) = 1*4*3 = 12 Numbers

Total = 20+20+20+12 = 72.
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by shovan85 » Mon Oct 04, 2010 5:21 am
shovan85 wrote:IMO A

Hundredth place either 8 or 9
Tenth place 0 to 9
Unit place 1,3,5,7,9

When Hundredth place is 8, Tenth place 0,2,4,6 (Not 8 as repeat), Unit place can be (1,3,5,7,9) = 1*4*5 = 20 Numbers
When Hundredth place is 8, Tenth place 1,3,5,7,9, Unit place can be (1,3,5,7,9 except the current tenth one) = 1*5*4 = 20 Numbers

When Hundredth place is 9, Tenth place 0,2,4,6,8, Unit place can be (1,3,5,7 Not 9) = 1*5*4 = 20 Numbers
When Hundredth place is 8, Tenth place 1,3,5,7(not 9), Unit place can be (1,3,5,7,9 except the current tenth one and 9) = 1*4*3 = 12 Numbers

Total = 20+20+20+12 = 72.
Typo last [color = red]When Hundredth place is 8[/color] is not 8 its 9.
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