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CEO
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How many numbers between 10,000 and 40,000 which **do not** [#permalink]
 06 Dec 2003, 16:27
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4.How many numbers between 10,000 and 40,000 which **do not** contain the digits 7 and 4 , can be divided by 15 ?
5.How many numbers between 10,000 and 100,000 **do not** have
reapeting digits and the the digits 3, 4 and 5 appear in ascending order(*but not necessarily adjacent*) ?
any NON - brute force ideas?
thanks
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Director
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for Q1. Numbers without 7 AND 4 we can have are 3x8x8x8x8=12 288 , numbers divisible by 15 should end with 5 OR 0, and the sum of numbers is divisible by 3, 10005,10010, 10015,10020,10025,10030,10035 so every third number is divisible by 15 or we have 12 288/3=4096...IMO 
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CEO
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BG wrote: for Q1. Numbers without 7 AND 4 we can have are 3x8x8x8x8=12 288 , numbers divisible by 15 should end with 5 OR 0, and the sum of numbers is divisible by 3, 10005,10010, 10015,10020,10025,10030,10035 so every third number is divisible by 15 or we have 12 288/3=4096...IMO need more help with => 3x8x8x8x8=12 288
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Director
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Since your number is between 10000 and 40000 first number we can select in 3 ways (1,2,3) the rest in 8 ways each ( without 7 and 4) which is ultimately 3x8x8x8x8
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CEO
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BG wrote: Since your number is between 10000 and 40000 first number we can select in 3 ways (1,2,3) the rest in 8 ways each ( without 7 and 4) which is ultimately 3x8x8x8x8
Wanna check your solution?
hint : the last digit can be only 5 or 0.
praetorian
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CEO
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BG wrote: is it 3072?
well ,3072/3 would be your "current" solution.
i am worried about your criteria for divisiblity ...
every third # divisible by 15??
are you sure it works for all the #'s
thanks
praetorian
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Q1.
Total No's are 3*8*8*8*2. For divisibility by 15, the Nos are
10,005, 10020, 10035 ... etc; so these are 15 appart. In this seris there will be no numbers with 7 or 4 as the digits. so we have 30,000/15 = 2,000 numbers.
It seesm that the first part about digits is redundant information. Is this solution correct. 
Last edited by bat_car on 21 Dec 2003, 10:11, edited 1 time in total.
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Director
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345XX as a pair XX can occupy 4 places in 3 of them we have 42 possibilities when is at the beginning only 36 possibilities( without 0)3x42=126+36=162 different numbers, when separate 12 ways of ordering the two numbers but in 6 of them it can not be 0 or 6x36+6x42=216+252=468, total 162+468=630 numbers..
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Intern
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praetorian123 could you please post the correct Ans.
Thanks
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