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How many positive multiples of 36 less than 100,000 can be

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How many positive multiples of 36 less than 100,000 can be [#permalink] New post 11 Aug 2006, 09:28
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How many positive multiples of 36 less than 100,000 can be formed by using only 6’s and 9’s?
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 [#permalink] New post 11 Aug 2006, 10:09
768.
If it's right, then I will explain later.
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 [#permalink] New post 11 Aug 2006, 11:17
Is it 6 ?

For a number to be multiple of 36 number must be divisible by 4 as well as 9.

So last digit will be 6 only.
Second last digit can not be 6 because 66 is not divisible by 4. So second last digit will be 9

1 digit numbers (None) = 0
2 digit numbers (96) = 1
3 digit numbers = 1
x96, x can be 9 only because sum of digits of 696 is not divisible by 9.
4 digit numbers (None) = 1
xx96 - Four possible values of xx - 66,69,96,99. Only 6696 has sum of digits divisible by 9.
5 digit numbers (None) = 3
xxx96 - Eight possible values for xxx
66696 - Sum of digits = 33
66996 - Sum of digits = 36 - YES
69696 - Sum of digits = 36 - YES
69996 - Sum of digits = 39
96696 - Sum of digits = 36 - YES
96996 - Sum of digits = 39
99696 - Sum of digits = 39
99996 - Sum of digits = 42

Total = 1+1+1+3 = 6
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 [#permalink] New post 11 Aug 2006, 12:39
there are altogather 2^4 numbers that are even because only 16 numbers with last digit 6 that are possibly divisible by 36. let me borrow dahiya's point that the last two digits must be 96. then the total possibilities are down to 2^3 = 8.

Among them i got only 6,696; 66,996; 69,696; and 96,696 divisible by 36. but i do not know the correct approach however i tried for coupple of minuets. i am curious to know.
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 [#permalink] New post 11 Aug 2006, 12:49
Combining dahiya's and professor's approach .
answer is: 4

Dahiya, there are no 2 digit and 3 digit nos. divisble by 36.

96 (sum = 15) not divisible by 9
696 (sum = 21)
996 (sum = 24)

Rest of the 4 no.s are what professor also found.
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 [#permalink] New post 11 Aug 2006, 12:53
Professor wrote:
there are altogather 2^4 numbers that are even because only 16 numbers with last digit 6 that are possibly divisible by 36. let me borrow dahiya's point that the last two digits must be 96. then the total possibilities are down to 2^3 = 8.

Among them i got only 6,696; 66,996; 69,696; and 96,696 divisible by 36. but i do not know the correct approach however i tried for coupple of minuets. i am curious to know.


i should edit my post. will do that later. i had to run :toilet therefore my post was incomplet.

i am curious to know the elegant approach.
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 [#permalink] New post 11 Aug 2006, 12:59
sgrover wrote:
Combining dahiya's and professor's approach .
answer is: 4

Dahiya, there are no 2 digit and 3 digit nos. divisble by 36.

96 (sum = 15) not divisible by 9
696 (sum = 21)
996 (sum = 24)

Rest of the 4 no.s are what professor also found.

These silly mistakes gonna cost me a lot. :beat :beat :wall :wall
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 [#permalink] New post 11 Aug 2006, 14:06
* there are 4 (or 2^2) four digit numbers that may be divisible by 36. among them only 6,696 is divisible by 36.

* there are 8 (or 2^3) five digit numbers that may be divisible by 36. Among them only 66,996; 69,696; and 96,696 are divisible by 36.

so altogather 4 numbers.
  [#permalink] 11 Aug 2006, 14:06
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How many positive multiples of 36 less than 100,000 can be

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