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# m25 - 20

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m25 - 20 [#permalink]

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30 Nov 2008, 06:20
How many odd three-digit integers greater than 800 are there such that all their digits are different?
a.40
b.56
c.72
d.81
e.104
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Re: m25 - 20 [#permalink]

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30 Aug 2009, 10:08
1
This post received
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For anybody like me who run's into this post looking for an answer, there's a later duplicated thread on this question. Solution is better explained there. Go to m25-76554.html
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Re: m25 - 20 [#permalink]

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30 Nov 2008, 06:56
Quote:
How many odd three-digit integers greater than 800 are there such that all their digits are different?
a.40
b.56
c.72
d.81
e.104

odd three-digit integers greater than 800 begin with 8 and 9, which are hundreds digits.
for 8: unit digits can be 1,3,5,7,9 - 5 numbers, and tens digits - 1,2,3,4,5,6,7,9 - 8 numbers (excepting 8, because all numbers should be different). 5x8=40 combinations for 8xx
for 9: units - 1,3,5,7 - 4 numbers, tens - 1,2,3,4,5,6,7,8, - 8 numbers (in this case 9 is out). 4x8=32

40+32=72, C
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Re: m25 - 20 [#permalink]

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30 Nov 2008, 07:27
mid number can also be 0, can't it?
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Re: m25 - 20 [#permalink]

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30 Nov 2008, 07:39
Quote:
mid number can also be 0, can't it?

you're right, I forgot about 0.
But after rethinking I've discovered that we have all the means 8 tens numbers for 8 and 8 tens numbers for 9 anyway: since units are 1,3,5,7,9 and 1,3,5,7 respectively, the tens numbers will be all of the 10 possible except fot those which are the same as units and hundreds ones, 10-2=8

for 8: 5 x 8 = 40
for 9: 4 x 8 = 32

answer is the same, 72, C
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Re: m25 - 20 [#permalink]

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30 Nov 2008, 07:47
Yes, it can. I got 72 for my answer.

801
803
805
807
809

There's 5 there and that's an even 10's digit

811 - can't count, two 1's
813
815
817
819

There are 4

For the 800's, when the ten's digit is even, there will be 5 numbers, and when the tens digit is odd, there will be 4, becuase one of the units digit will be the same as the tens digit and cannot be counted....so there will be

5 sets of 5 and 5 sets of 4 (5 sets of even ten's digits and 5 sets of odd ten's digits) so 5*4 + 5*5 = 45

Now for the 900's. The same concept as the 800s will apply, but instead of 5 for the evens and 4 for the odd tens, we'll have 4 and 3 because whenever we have a 9 in the units, this will be the same as the hundreds and cannot be counted.

so 5 sets * 4 + 5 sets * 3 = 20 + 15 = 35. 45 + 35 = 80...but in the 800s, we can't count any from the 880's and in the 900's we cant count any from the 990s. So the 880's is an even tens, so there would be 5 we counted that we should not have and the 990s is an odd, so there would be 3 we counted (991, 993, 995, 997. We already accounted for 999 because the units & tens digit were the same). 5 + 3 = 8 too many. 80 - 8 = 72.

This might not be the fastest way of going about it, but I believe it can be worked in under 3 minutes. It's a simple question and with familiarity from practice problems like this, the time on test day can be much less.

Answer: 72

botirvoy wrote:
mid number can also be 0, can't it?

botirvoy wrote:
mid number can also be 0, can't it?

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Director
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Re: m25 - 20 [#permalink]

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30 Nov 2008, 10:57
Having seen how you guys approached it, I worked out the following:

Between 800 and 899:
mid digit even: 1x4x5=20 - i.e. for hundredth -8; for tenth - 0,2,4,6; for unit - 1,3,5,7,9
mid digit odd: 1x4x5=20

Between 900 and 999:
mid digit even: 1x5x4=20
mid digit odd: 1x3x4=12

72 odd three digit integers with different digits.
Would be keen to see other ways of solving it.
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Re: m25 - 20 [#permalink]

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30 Aug 2009, 01:37
Please help.

I solved this one using jallenmorris' approach. Should have been easy, but wasn't. I mistakenly accounted for 881,883,885,887,889, so I was getting 77 as the answer, and I wasted a lot of time on this question.

Now, I don't understand how the faster approach (8x5+8x4) can be correct.

Lets take 8 as the hundreds digit. Solution proposed says that we have 8 possibilities for the tens digit (1-2-3-4-5-6-7-9) and 5 possibilities for the units digit (1-3-5-7-9) ... therefore, 8x5 gives all the combinations.

But the way I see it, 8x5 gives ALL the combinations, including the unwanted ones 811, 833, 855, 877 and 899. So how can this be correct?

Could somebody help me in finding my reasoning flaw?
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Re: m25 - 20 [#permalink]

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30 Aug 2009, 10:42
I go with ans: 72 as well

since number between 800 and 999, we can be split into two sets.

1) from 800 to 899

8 _ _

The units digit can be filled with any of the five odd numbers (1,3,5,7,9) . Now since two numbers are used (8 and one odd number in the unit's digit), we can fill the ten's place with the remaining 8 numbers (left over digits = 10 - 2).
hence the total possible options is = 8 * 5 = 40

2) from 900 to 999

9 _ _

The units digit can be filled with only four of the five odd numbers (1,3,5,7) since all digits need to be different. And since two numbers already used up , the ten's place can be filled with the left over 8 numbers .
hence the total possible options is = 8 * 4 = 32

adding - 40 + 32 = 72
Re: m25 - 20   [#permalink] 30 Aug 2009, 10:42
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# m25 - 20

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