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The largest possible 5-digit number is 99999 Divide 99999 by 6, get 16666.5. So the largest integer that does not exceed that is 16666. So we multiply 16666 by 6, getting 99996. Edwin
$$\color{red}{416/3125}=0.13312. $$ 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 $(2/5)\cdot(2^4+4\cdot2+12\cdot2^3+4\cdot2^4+6\cdot2^2)/5^4$.
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Given:
The greatest number divisible by 6, 7, 8, and 12.
Concept used:
LCM method used
Calculation:
Now finding the LCM,
6 = 3 × 2
7 = 7 × 1
8 = 2 × 2 × 2
12 = 2 × 2 × 3
So, required LCM = 2× 2 × 2 × 3 × 7 = 168
Therefore, the number has to be of form 168 x + 4, where x is an integer.
The largest number of 5 digits is 99999
When divided by 168, 99999 leaves the remainder 39, which is
99999 = 168 × 595 + 39
So, 99999 - 39 = 99960 which is divisible by 168.
Now as it leaves a remainder 4 in each case
Therefore, the required number is 99960 + 4 = 99964
∴ The correct answer is 99964.
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