The main constraint is that each digit can only take digits from $\{1, 2, 3, 4, 5\}$. So the sample space will be $5^{5}$.

What is the probability that a random number taken from this sample space will be divisible by $6$?


  • 2
    $\begingroup$ If you wanted a rough guess, I'd go with 1/6. If you want a better guess, a number is divisible by 6 if and only if it is divisible by 2 and by 3. 2 is easy because that means the number needs to end in 2 or 4. 3 means the digits have to add up to a multiple of 3. So, count the number of numbers satisfying these contraints. $\endgroup$ – GeoffDS Feb 17 '12 at 11:52
  • 1
    $\begingroup$ Only two-fifths of the numbers will be even, so a slightly less rough rough guess would be two-fifteenths. $\endgroup$ – Gerry Myerson Feb 17 '12 at 12:10
  • $\begingroup$ @Graphth: Counting such numbers is the biggest challenge. If u can give a method for the counting even in a rough manner, it would be great. $\endgroup$ – Maverickgugu Feb 17 '12 at 12:19
  • $\begingroup$ As you can see, @Gerry's estimate agrees with Didier's answer to within 0.2%. $\endgroup$ – Willie Wong Feb 17 '12 at 15:38
  • $\begingroup$ @Willie, another way to put it is that two-fifteenths of 3125 is 416-and-two-thirds, while Didier's count is 416. $\endgroup$ – Gerry Myerson Feb 17 '12 at 22:38

$$\color{red}{416/3125}=0.13312. $$ The last digit must be $2$ or $4$, this happens with probability $2/5$. The sum of the four other digits must be $\pm1\pmod{3}$, according to the last digit being $2$ or $4$. Since both events have the same probability, the answer is $2/5$ times the probability that the sum $s$ of four digits is $1\pmod{3}$, that is, $s=-2$ or $s=+1$ or $s=+4$.

$s=+4$ corresponds to $+1,+1,+1,+1$, with probability $2^4/5^4$.

$s=+1$ corresponds to $0,0,0,+1$, or $0,+1,+1,-1$ in whatever order. In the first case, one must place the $+1$, thus $4$ cases, with probability $2/5^4$ each. In the second case, one must place the $0$ and the $-1$, thus $12$ cases, with probability $2^3/5^4$ each.

$s=-2$ corresponds to $+1,-1,-1,-1$, thus $4$ cases, with probability $2^4/5^4$ each, or to $0,0,-1,-1$, thus $6$ cases, with probability $2^2/5^4$ each.

Summing up, the answer is $(2/5)\cdot(2^4+4\cdot2+12\cdot2^3+4\cdot2^4+6\cdot2^2)/5^4$.

| cite | improve this answer | |
  • $\begingroup$ that was a neat one. :) $\endgroup$ – Maverickgugu Feb 17 '12 at 13:14

That probability is 1/6-P. Where P is the probability of a number be ( in this range) lesser than the first (minor) 6 multiple (in this range) or greater than the last (major) 6s+5 number (in the range).

Evaluate P.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.