Take the number 3336. You know it's divisible by 1 because everything is. You know it's divisible by 2 because it's even. You know it's divisible by 3 because the digits add up to a multiple of 3 and you know it's divisible by 4 because the last two digits are divisible by 4. So you've found at least four factors: 1,2,3 and 4.
If a number is divisible by both three and four, it's divisible by twelve.
If it's to be divisible by 5, it must end in 5 or 0. And if it's to be divisible by 3, adding the digits together must result in a multiple of 3.While some numbers that end in zero will still work, you can guarantee that the number won't be divisible by four by choosing a number that ends in five.So you'd need a four-digit number, that ends in 5, whose digits add up to a multiple of 3. The smallest such number would be 1005, and the largest would be 9975.(However, the actual largest number that would fit your question is 9990; remember that I did say some numbers that end in zero would still work.)
18, 24, 32, 36
No. Four is not evenly divisible by 25.
9996
It is 9996
1008 is the lowest four-digit number divisible by 36, and 9972 is the highest.
The lowest 4-digit number that is divisible by 6 is 1002
To find a four-digit number divisible by both 8 and 9, we need to find the least common multiple (LCM) of 8 and 9, which is 72. The smallest four-digit number divisible by 72 is 1008 (72 x 14 = 1008).
Greatest = 9,999Smallest = 1,008
1002 is the smallest 4 digit number divisible by 3 and 2.
The integers divisible by those 5 numbers are exactly the multiples of 1008. The largest multiple of 1008 which is still a four-digit number is 9*1008 = 9072.
fcfdfd
1100
They all are.
3333