What is the probability that a two digit number picked up at random will be divisible by 5

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"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.

1. 6/13
2. 8/15
3. 2/5
4. 3/5
5. None of these

GIVEN:
The number should be multiple of 4 and 5

CONCEPT:

For a number to be a multiple of two numbers it should be the multiple of their L.C.M

FORMULA USED:
P = Favorable outcomes/Total outcomes

CALCULATION:
Total 2 digit numbers = 90 (from 10 to 99)

The number should be divisible by 20 (L.C.M of 4 and 5)

2 digit numbers divisible by 20 are, 20, 40, 60, 80

Required probability = 4/90 = 2/45

∴ The probability is 2/45. 

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