The probability that a two digit number is a multiple of 3 and not a multiple of 5

"Either $d$ or $e$ has to be $6$", this is incorrect. In order for the number $de$ to be divisible by $3$, the sum of the two digits $(d+e)$ should be a multiple of $3$. Therefore, $(d+e)=(4+5)$ or $(5+7)$

Proof: Suppose you have a three-digit number $abc$ Then$$abc=100a+10b+c=99a+9b+0c+(a+b+c)$$ if $abc$ is divisible by $3$, then $99a+9b+0c+(a+b+c)$ must also be divisible by 3. Now because $99a,9b,0c$ can clearly be divided by $3$ (because of $99$), so $(a+b+c)$ must be a multiple by $3$.

For higher-digit numbers this proof also holds.

IBM Company Numerical Ability Probability

Total two digit numbers = 90 The numbers multiple of 3 but not 5 = 12,18,21,24,27,33,36,39,42,48,51,54,57,63,66,69,72,78,81,84,87,93,96,99 = 24 numbers
Hence probability = 24/90 = 4/15
10 years agoHelpfull: Yes(37) No(0)
Since every third number starting from 10 will be divisible by 3, so total number of numbers divisible by 3 are 90/3 = 30 Numbers which are divisible by 3 and 5 both are numbers which are multiple of 15. For the range 10 to 99, 15 is the first number divisible by 15 and 90 is the last number. So total number of numbers divisible by 15 are: (90-15)/15 + 1 = 5 + 1 = 6 Number of numbers which are divisible by 3 are 30 and number of numbers which are divisible by 3 and 5 both are 6. So number of numbers divisible by 3 and not by 5 are: 30-6=24 So total probability = 24/90 = 4/15
8 years agoHelpfull: Yes(3) No(0)
total 2 digit numbers from 1 to 100 is 90 the 2 digit numbers multiplied by 3 is 12,15,18,21,24.....99=30numbers
so the probability is 30/90=1/3
10 years agoHelpfull: Yes(2) No(18)
10-30 number mul of 3 is 7 (15 il be div by 5 )so 7-1=6 31-60 =10 (45..) 10-1 = 9 61-90= 9 93 96 99 = 3
3+9+9+6 = 27
10 years agoHelpfull: Yes(1) No(5)
total 2 digit nos=90 multiples of 3 which are two digit=29 multiples of 5 b/w these=6 so 29-6=23
hence total prob= 23/90!!!
10 years agoHelpfull: Yes(1) No(6)