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bloody tough no. prop sum

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by xcusemeplz2009 » Mon Oct 19, 2009 3:04 am
total digit =10...(0,1,2,3,....9)
total odd digit=5...(1,3,5,7,9)
let no be xyz
two cases first x can be 8 or 9 ( as no is > 800)

first case :fix x as 8(1 way)
z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways
total ways=1*5*8=40

second case : fix x as 9=> 1 way
z can be selected out of 4 odd digit( 9 alraedy selected) in 4 ways
Y can be seleceted from 8 digits in 8 ways

total=1*8*4=32

total possible ways =40+32=72
It does not matter how many times you get knocked down , but how many times you get up
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by vkb16 » Mon Oct 19, 2009 4:12 am
z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways
why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??
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by BuckeyeT » Mon Oct 19, 2009 6:25 am
vkb16 wrote:
z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways
why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??
"How many odd three-digit integers..."

By his formula, xyz, z must be odd. So, it can only be selected out of odd integers. If it was "2", an even integer, the number would be xy2. And for all values, (xy), xy2 would be even. For example, 842 is even. Since we're only looking for odd numbers, z must be selected as an odd number.
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by xcusemeplz2009 » Mon Oct 19, 2009 8:22 am
vkb16 wrote:
z have any one value out of odd digits therefore 5 ways to select
now for y digit left 8 so ways
why must be selected only out of odd digits, whereas Y can be selected out of all remaining numbers??
the question is asking for a three digit odd no. and for any no. to be odd the digit at one's place has to be a odd
It does not matter how many times you get knocked down , but how many times you get up
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by vkb16 » Tue Oct 20, 2009 12:00 am
thanks a lot guys! that made it clear...
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