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shabbir 17Aug2009 17:43

Number Format | 17 Aug 2009
 
Find the number of the form

X1X2X3X4X5X6X7X8
  • X1 to X8 are all unique single digit numbers
  • X1 to X8 none of them is zero
  • X1 to X8 none of the is 9
  • X1X2 is divisible by 8
  • X2X3 is divisible by 7
  • X3X4 is divisible by 6
  • X4X5 is divisible by 5
  • X5X6 is divisible by 4
  • X6X7 is divisible by 3
  • X7X8 is divisible by 2

I hope I had the coffee correctly today

SaswatPadhi 17Aug2009 17:54

Re: Number Format | 17 Aug 2009
 
72185634 :d

nimesh 17Aug2009 18:18

Re: Number Format | 17 Aug 2009
 
Nice question and nice reply too :)

hoping this is correct, Congrats Saswat ;)

SaswatPadhi 17Aug2009 18:20

Re: Number Format | 17 Aug 2009
 
Explanation :

Let us consider (for clarity), the number to be PQRSTUVW.

So, ST must be divisible by 5.
--> As no zero is allowed only possibility is T = 5.
--> U must be 2 or 6 as TU is divisible by 4.

Now, PQ must be divisible by 8, and QR by 7.
--> PQ can be 16, 24, 32, 48, 64, 72.
--> QR can be 14, 21, 28, 42, 63.
--> Clearly, PQR can be 163 or 321 or 328 or 721 or 728.
--> Corresponding possible values of PQRS for above PQR values are : 1636, 3218, 3284, 7218, 7284.
--> Clearly 1636 is impossible, 'coz of repetition of 6. So, we are left with the following values of PQRSTU yet : 321856, 328456, 721856, 728456.

Now, UV is divisible by 3. As, U can only be 6 (as seen from above list), only 63 is the possible case.
So, we can reduce the possible set to : 7218563, 7284563 now.
The only thing we are to do now is to make the number even, by adding a digit at the right. We can add only 2, 4, 8 or 6.
In the first case, we can use only 4, 'coz all others are used up.
In the second, we cannot use any of them, so it can be rejected.

We are finally, left with 72185634.

shabbir 17Aug2009 19:20

Re: Number Format | 17 Aug 2009
 
Great

SaswatPadhi 17Aug2009 19:31

Re: Number Format | 17 Aug 2009
 
Thanx :)

mayjune 17Aug2009 19:42

Re: Number Format | 17 Aug 2009
 
well done saswat...


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