2, 5 or 8.
If the number is divisible by 9 the sum of its digits (its digital root) must be divisible by 9. 5+6+8+9 = 28 The next multiple of 9 is 36 and that is 8 more. So the missing digit must be 8.
shut up and do something better with your life
No single digit determines divisibility by 3.
A solution.
342 is divisible by 6 and not 5.Numbers divisible by 5 end in zeros or fivesNumbers divisible by 6 have a slightly more complicated rule. If, when the individual digits are added together and they make a multiple of 3, see if the number is even. If so, and the digits add up to a multiple of 3, then the number is a divisible by 6.
If the number is divisible by 9 the sum of its digits (its digital root) must be divisible by 9. 5+6+8+9 = 28 The next multiple of 9 is 36 and that is 8 more. So the missing digit must be 8.
0.
shut up and do something better with your life
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.
Any number divisible by 3 has digits that add up to a multiple of 3. Adding a 1 to any of these digits or inserting a 1 anywhere in the number would make it divisible by 3.
No single digit determines divisibility by 3.
In mathematical logic, An integer A if divisible by 100 iff the last two digits are 0. "iff" stands for "if and only if".
A solution.
The 5 should be deleted.
342 is divisible by 6 and not 5.Numbers divisible by 5 end in zeros or fivesNumbers divisible by 6 have a slightly more complicated rule. If, when the individual digits are added together and they make a multiple of 3, see if the number is even. If so, and the digits add up to a multiple of 3, then the number is a divisible by 6.
A number is divisible by 8 if and only if it is an integer and the last three digits make a number that is divisible by 8. A general divisibility rule for 2^n where n is an integer is to take the last n digits and if they make a number that is divisible by n then it is divisible by n.
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.