Itself because 345/345 = 1
All multiples of 345, which is an infinite number.
No. Numbers that are divisible by 2 must end in an even number (0,2,4,6,8)
123 + 345 = 468
468
234
For each given number, add the digits. If the sum comes to nine, then it is divisible by '9'. 234 ; 2 + 3 + 4 = 9 So is dividble by '9' 345 ; 3 + 4 + 5 = 12 = 1 + 2 = 3 Not divisible by '9' 567 ; 5 + 6 + 7 = 18 = 1 + 8 = 9 so is divisible by '9'.
Itself because 345/345 = 1
All multiples of 345, which is an infinite number.
No. Numbers that are divisible by 2 must end in an even number (0,2,4,6,8)
234 x 345 = 80,730
123 + 345 = 468
468
The number half way between them was 345.
1, 3, 5, 15, 23, 69, 115, 345.
No, it is not divisible by 9. To check if a number is divisible by 9 add all the digits together and if this sum is divisible by 9, then so is the original number. As the check can be applied to the sum, keep adding the digits together until a single digit remains. If this digit is 9, then the original number is divisible by 9 (otherwise it gives the remainder when the original number is divided by 9). for 345: 3 + 4 +5 = 12 for 12: 1 + 2 = 3 which is not 9, so 345 is not divisible by 9. (The remainder when 345 is divided by 9 is 3.)
12 345