ronr34 wrote:
I tried to do as follows:
take all 5 digit numbers possible : 5 *5*4*3*2
divide by 3 to get all numbers divisible by 3.
What is wrong with this logic?
We cannot do this because we have the asymmetric 0 as one of the digits. The number of 5 digit numbers that can be formed with 0, 1, 2, 3 and 4 is different from the number of 5 digit numbers that can be formed with 1, 2, 3, 4 and 5 (because 0 cannot be the first digit).
Had the digits been 1, 2, 3, 4, 5 and 6, then your method would have been correct.
If 0 is included:
{0, 1, 2, 3, 4} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 3, 5} --> 96 5-digit numbers possible with this set.
{0, 1, 2, 4, 5} --> 96 5-digit numbers possible with this set. - All these numbers are divisible by 3
{0, 1, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{0, 2, 3, 4, 5} --> 96 5-digit numbers possible with this set.
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
The number of 5 digit numbers in these sets is not the same - Sets with 0 have fewer numbers
If 0 is not included:
{1, 2, 3, 4, 5} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 4, 6} --> 120 5-digit numbers possible with this set.
{1, 2, 3, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
{1, 2, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{1, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set.
{2, 3, 4, 5, 6} --> 120 5-digit numbers possible with this set. - All these numbers are divisible by 3
Here exactly 1/3rd of the numbers will be divisible by 3.
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Karishma
Veritas Prep GMAT Instructor
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