4: to make it 90144 = 15024*6
99996
To determine what digit can be placed in the blank to make a number divisible by 3, you need to ensure that the sum of all the digits in the number, including the blank, is divisible by 3. For example, if the number is 25_, you would calculate the sum of 2 + 5 + _ and check which digit (0-9) makes the total divisible by 3. Once you find the appropriate digit, you can fill in the blank.
To make a number divisible by 10, its last digit must be 0. The last digit of 23483 is 3. Therefore, to make it divisible by 10, you should subtract 3 from 23483. This means the least number that should be subtracted is 3.
The 3, to the left.
If the number is even, it is a multiple of 2 If the sum of the digits make a number divisible by 3, the number is a multiple of 3 If the number ends in 5 or 0, the number is a multiple of 5 If the number is divisible by 2 and 3, the number is a multiple of 6 If the sum of the digits make a number divisible by 9, the number is a multiple of 9
99996
To determine what digit can be placed in the blank to make a number divisible by 3, you need to ensure that the sum of all the digits in the number, including the blank, is divisible by 3. For example, if the number is 25_, you would calculate the sum of 2 + 5 + _ and check which digit (0-9) makes the total divisible by 3. Once you find the appropriate digit, you can fill in the blank.
To make a number divisible by 10, its last digit must be 0. The last digit of 23483 is 3. Therefore, to make it divisible by 10, you should subtract 3 from 23483. This means the least number that should be subtracted is 3.
123456
6 (or 0)
The 3, to the left.
If the number is even, it is a multiple of 2 If the sum of the digits make a number divisible by 3, the number is a multiple of 3 If the number ends in 5 or 0, the number is a multiple of 5 If the number is divisible by 2 and 3, the number is a multiple of 6 If the sum of the digits make a number divisible by 9, the number is a multiple of 9
29 is not divisible by 3 and how any digit will alter that depends on how that digit is to interact with 29.
Any integer divisible by 9 has digits whose sum is divisible by 9. The sum of the digits in 2530 is 2+5+3+0 = 10, and the sum of 1+0 = 1. Having an additional digit "8" in the number, or replacing the "0" with an "8", would make a number divisible by 9. Thus 82,530; 28,530; 25,830; 25,380; 25,308; or 2,538 are all divisible by 9.
There is no digit that can replace the ? in 2?67 to make it divisible by 4. All multiples of 4 are even, 2?67 is an odd number and so cannot be a multiple of 4.
I don't completely understand the question, but there's a trick: A number is divisible by 3 if and only if the sum of its digits is divisible by 3. This should guide you to an answer.
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