The smallest one would be 10.
Since the LCM of 2, 5, and 10 is 10, then all multiples of 10 are divisible by 2, 5, and 10.
There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.
If its divisible by 5 AND 2 it must be divisible by 10 So you just have to pick the only number between 21 and 39 that's divisible by 10
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.
10
A number that is divisible by 2, 3, 4, 5, and 10 must be divisible by the least common multiple of these numbers, which is 60. Therefore, any number that is a multiple of 60 will be divisible by 2, 3, 4, 5, and 10. This is because 60 is the smallest number that contains all the prime factors of 2, 3, 4, 5, and 10.
Because 10 is divisible by both 2 and 5
10 is divisible by both 5 and 2, as are any multiples of 10.
There are no numbers that satisfy this. If a number is divisible by both 2 and 5, then it must also be divisible by 10.
If a number is divisible by 10 (i.e. 100) then it is also divisible by 5. Ex: 100 by 10, is 10 by 5, is 20 20 by 10 is 2 by 5 is 4 The number of times it is divisible by 5 will always be double the number of times it is divisible by 10, which makes sense because 10 is double 5.
It is divisible by 2, 5 and 10 but not the rest.
Always. A number divisible by another is also divisible by all the factors of that number. A number that is divisible by 10 is also divisible by 2 and 5. Take any multiple of 10 and check it. 70/5=14 or 180/5=36
If its divisible by 5 AND 2 it must be divisible by 10 So you just have to pick the only number between 21 and 39 that's divisible by 10
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.
Every number divisible by 10 is divisible by 5.
No. (Example.....10 / 2 = 5, but 10 / 4 = 2.5)
A number divisible by both 2 and 5 will be divisible by their product (2 x 5), which is 10. Any number divisible by 10 ends in 0. The only number listed that ends in 0 is 110.
Yes.