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@morningdot-6Xuj4M • Aug 4, 2010
On the first one 0,1,2,3,5,7 and on the 2nd one 0,1,2,4,6,8.
Am i correct?? -
@vinsocorp-4fscJ8 • Aug 5, 2010
Mohit its a good try. think how u show 9,19,29.one clue: one digit number u can show it by single cube,another cube is eliminated.
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@silverscorpion-iJKtdQ • Aug 5, 2010
Well, there are atmost 31 days in a month. So, till 31, there are 2 numbers which have repeated digits - 11 and 22.
So, both the cubes must have 1 and 2, or they cant display 11 and 22.
So, there ia 1 and 2 in both cubes. There are 8 digits remaining (3 4 5 6 7 8 9 0)
There are eight sides remaining in the two cubes, totally..
Hence, the remaining 8 digits can be filled in any manner in the remaining sides..
For example,
1 2 3 4 5 6 for the first cube,
1 2 7 8 9 0 for the second is a possibility.
Or, 1 2 4 6 8 0 for first,
1 2 3 5 7 9 for second
etc.. Any combination of the above will be fine, as long as both cubes have 1 and 2.. -
@morningdot-6Xuj4M • Aug 5, 2010
vinsocorpMohit its a good try. think how u show 9,19,29.
For 9
0 and reverse of 6
for 19
1 and reverse of 6
for 29
2 and reverse of 6.
.....
vinsocorpone clue: one digit number u can show it by single cube,another cube is eliminated.
You doesn't mentioned this in question.
And i tell you that it's a vary simle question when you alow for the above quotes.
To make a question standard you have to reduce the probablity of finding a answer. -
@vinsocorp-4fscJ8 • Aug 7, 2010
Mohit u r very well in solving puzzles.here after i maintain standards in my question. ......what u doing...
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@marustech-CKSJnz • Aug 7, 2010
very good .. but how we get 7,8,9 separateiy
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@marustech-CKSJnz • Aug 7, 2010
great thinking
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@learner-SrHIlM • Aug 7, 2010
we can also do this sequence
1st cube: 1,2,3,6,4 2nd cube: 0,1,2,5,7,8 -
@saandeep-sreerambatla-hWHU1M • Aug 20, 2010
silverscorpionWell, there are atmost 31 days in a month. So, till 31, there are 2 numbers which have repeated digits - 11 and 22.
So, both the cubes must have 1 and 2, or they cant display 11 and 22.
So, there ia 1 and 2 in both cubes. There are 8 digits remaining (3 4 5 6 7 8 9 0)
There are eight sides remaining in the two cubes, totally..
Hence, the remaining 8 digits can be filled in any manner in the remaining sides..
For example,
1 2 3 4 5 6 for the first cube,
1 2 7 8 9 0 for the second is a possibility.
Or, 1 2 4 6 8 0 for first,
1 2 3 5 7 9 for second
etc.. Any combination of the above will be fine, as long as both cubes have 1 and 2..
This is the straight forward simple answer!!
๐