1002 is the smallest 4 digit number divisible by 3 and 2.
9996
There is no answer because for a number to be divisible by 8, it would end in an even number and therefore be divisible by 2.
There are 18 numbers with 2 digits that are divisible by 5. First 2 digit number is 10 → 10 ÷ 5 = 2 → first 2 digit number divisible by 25 is 5 × 2 Last 2 digit number is 99 → 99 ÷ 5 = 19 4/5 → last 2 digit number divisible by 5 is 5 × 19 → There are 19 - 2 + 1 = 18 numbers with 2 digit divisible by 5.
I think 2520
1200 is one such number.
12 is the smallest 2 digit number divisible evenly by 4
There are 18 numbers with 2 digits that are divisible by 5. First 2 digit number is 10 → 10 ÷ 5 = 2 → first 2 digit number divisible by 25 is 5 × 2 Last 2 digit number is 99 → 99 ÷ 5 = 19 4/5 → last 2 digit number divisible by 5 is 5 × 19 → There are 19 - 2 + 1 = 18 numbers with 2 digit divisible by 5.
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.
Yes. If the last digit is even (ie one of 0, 2, 4, 6, 8) then the whole number is divisible by 2. The last digit is 4, so the number is divisible by 2.
A number which which ends with a 2 digit number that is divisible by 4. For example, 1028 ends with 28 which is divisible by 4.
1800
There is no number divisible by zero.
These are simple ones , Probably not good ones but i'll give it a go. It is the 4th number, it can be divided by 2 AND.. its four units away from zero. ----------------------------------------------- It is divisible by 2 and 1 It is not a prime number It is an even number It is a single digit number
Numbers that are co-prime won't both be divisible by nine.
9990
By 3 only.It is not divisible by 2, 4, 5, 6, 9 & 10.Division tests:2:If the number is even, ie if the last digit is even (0, 2, 4, 6, 8) it is divisible by 2. 633 is odd (last digit is 3 which is not an even digit), so 633 is not divisible by 2.3:Sum the digits; if the sum is divisible by 3, the original number is divisible by 3. (Can repeat the summing until a single digit remains; if this digit is 3, 6 or 9 (ie divisible by 3) then the original number is divisible by 3.)6 + 6 + 3 = 151 + 5 = 66 is divisible by 3, so 663 is divisible by 3.4:Add the last (units) digit to twice the previous (tens) digit; if this sum is divisible by 4, so is the original number. (Can repeat summing until a single digit remains; if this digit is 4 or 8 (ie divisible by 4) then the original number is divisible by 4.)6 x 2 + 3 = 151 x 2 + 5 = 77 is not divisible by 4, so 663 is not divisible by 4.Note: all multiples of 4 are even; 633 is odd so it cannot be divisible by 4 (the above test does not need to be done).5:The last digit of the number must be 0 or 5. 3 is not 0 nor 5, so 663 is not divisible by 5.6:Number must pass the 2 and 3 tests (above). Fails 2 test (above), so 663 is not divisible by 6.9:Sum the digits; if the sum is divisible by 9, the original number is divisible by 9 (Can repeat the summing until a single digit remains; if this digit is 9 (ie divisible by 9) then the original number is divisible by 9)6 + 6 + 3 = 151 + 5 = 66 is not 9, so 663 is not divisible by 9.10:Last digit must be 0. 3 is not 0, so 663 is not divisible by10.