Complete the number so that it is divisible by 9 without a remainder:
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Complete the number so that it is divisible by 9 without a remainder:
The problem asks us to find the digit represented by '?' in the number so that the entire number is divisible by 9.
To solve this problem, we will use the divisibility rule for 9, which states: A number is divisible by 9 if the sum of its digits is divisible by 9.
Let's calculate the sum of the known digits:
Now let's include the unknown digit, represented by , in the sum. The total sum of the digits will be .
We need to be divisible by 9. So, find a value for such that .
Let's check each value for from 0 to 9:
Thus, the sum of the digits is divisible by 9 when .
The correct digit to complete the number such that it is divisible by 9 is .
Determine if the following number is divisible by 3:
\( 564 \)
This works because of how our number system is built! When you write a number like 427, it's really . The magic is that 100, 10, and 1 all leave remainder 1 when divided by 9.
For single missing digits, you'll always get exactly one answer! Since we only check digits 0-9, and only one will make the sum divisible by 9, there's no ambiguity.
Not always! You can be smarter: if your known sum is 13, you need to add something to reach the next multiple of 9. Since , you know the answer is 5 right away!
Yes! The digit sum rule works for 3 and 9. For 6, a number must be even AND divisible by 3. But for other numbers like 7 or 11, you need different tricks.
Then the missing digit would be 0! For example, if your known digits sum to 9 or 18, adding 0 keeps it divisible by 9.
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