For a number to be divisible by 6, it must satisfy two conditions:

1. Divisibility by 2: The last digit must be even.
2. Divisibility by 3: The sum of the digits must be divisible by 3.

Let's apply these rules to the number: $495296396d856776372$.

1. Condition for Divisibility by 2:

• The last digit is 2, which is even. Therefore, the number satisfies the condition for divisibility by 2.

2. Condition for Divisibility by 3:

• We need to ensure the sum of all digits (including $d$) is divisible by 3.

Sum of digits (excluding $d$):

$4 + 9 + 5 + 2 + 9 + 6 + 3 + 9 + 6 + 8 + 5 + 6 + 7 + 7 + 6 + 3 + 7 + 2 = 104$

Now, to make the number divisible by 3, the total sum $104 + d$ must be divisible by 3. We calculate $104 \mod 3$:

$104 \div 3 = 34 \text{ remainder } 2$ This means $104 \equiv 2 \mod 3$. For the total sum to be divisible by 3, we need $104 + d \equiv 0 \mod 3$. So, $d \equiv 1 \mod 3$.

Thus, $d$ can be any digit that leaves a remainder of 1 when divided by 3. The possible values for $d$ are:

$d = 1, 4, 7$

Conclusion:

The digits that $d$ can be replaced with to make the number divisible by 6 are 1, 4, or 7.

Would you like further explanation on any part of this solution?

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Here are 5 related questions for further exploration:

1. How is the divisibility rule for 3 derived?
2. Can you apply the same method to check divisibility by 9?
3. What if the number ends with an odd digit—what does that mean for divisibility by 6?
4. How would you determine the smallest number divisible by both 4 and 6?
5. What changes if you consider the divisibility rule for 12 instead of 6?

Tip: Divisibility rules are quick shortcuts to determine if a number is divisible by a specific integer without performing division!