2400
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
9990 is one such number.
Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...
There is no 4-digit number that is divisible by 2356 and 10.
There is no number that is "divisible by 4" and "not divisible by 4" at the same time - a number cannot be both a multiple of 4 and not a multiple of 4.
2400
A 4-digit number divisible by both 5 and 9 must be divisible by their least common multiple, which is 45. To find a 4-digit number divisible by 45, we need to find a number that ends in 0 and is divisible by 45. The smallest 4-digit number that fits these criteria is 1005 (45 x 22 = 990, and adding 15 gives us 1005).
9990 is one such number.
3000 is one possibility
If the last two digits are divisible by 4 then the number is divisible by 4. Thus, if the tens digit is even and the units digit is 0 or 4 or 8 OR if the tens digit is odd and the units digit is 2 or 6 then the number is divisible by 4.
Any number whose last digit is 0 would be divisible by both these numbers. 1000, 1010, 1020...
There is no 4-digit number that is divisible by 2356 and 10.
smallest 4-digit number divisible by 2 = 1000
To be divisible by both 4 and 9 a number must be a multiple of the Least Common Factor of 4 and 9. The LCM is 36. The first 3 digit number in each hundred that is a multiple of 36 are as follows :- 108, 216, 324, 432, 504, 612, 720, 828 and 900.
There is no limit to the number of digits.If, for example, a X is a k-digit number which is divisible by 4 then 10*X is divisible by 4 and 10*X will be a (k+1)-digit number.
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