Every number divisible by 10 is divisible by 5.
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.
Any number that ends in a zero is divisible evenly by 5 and 10
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 a number ends with 0, it is divisible by 10. If a number ends with 5 or 0, it is divisible by 5.
Three is a prime number and isn't divisible by any whole number * * * * * True, but irrelevant to the question. Any number that is divisible by 10 MUST be divisible by 5. Therefore there are no such numbers.
A prime number is a number which is divisible by 1 and itself. The numbers from 1 to 10 which meet this criteria are 2, 3, 5, and 7
no, it is divisible by 1, 2, 5, 7, 10, 14, 35, 70
155 is not a prime number, but a composite number, as it is divisible by 5 and 31.
No anything that ends with a zero or a five is divisible by five but anyhting with a decimal IS NOT
To find a number between 80 and 100 that is divisible by both 5 and 6, we need to find the least common multiple (LCM) of 5 and 6, which is 30. We then look for a multiple of 30 between 80 and 100, which is 90. Therefore, the number 90 between 80 and 100 is divisible by both 5 and 6.
Construct DFA for the language
No. For one thing, 25 is an odd number (ends in 5). 10 is even, meaning is it divisible by 2. All even numbers are divisible by 2; all numbers divisible by 2 are even. 25 isn't one of them. So it is not a multiple of 10. For another, just try multiplying 10, the way you would count dimes or ten-dollar bills: 10, 20, 30...did you mention 25? 25 is two and a half tens.
403 is divisible by no number
The number is only divisible by 1 and itself. Examples 1, 3, 5, 7, 11
Any number that ends in a zero is divisible by 10. In general, any power of ten is a multiple of ten. 102 = 10 x 10 = 100 103 = 10 x 10 x 10 = 1,000 104 = 10 x 10 x 10 x 10 = 10,000 Any one (1) with some zeros after it is a multiple of ten.
To find the smallest number that has 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 as factors, you need to find the least common multiple (LCM) of these numbers. The LCM is the smallest number that is divisible by all the given numbers. The LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 is 2520. Therefore, 2520 is the smallest number that has 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10 as factors.
The divisibility rule for 3 is as follows: If the sum of the digits of a number is divisible by 3 then the number is divisible by 3. For example, the number 21 is divisible by 3 since 2+1=3, and 3 is divisible by 3. The number 49 is not divisible by 3 since 4+9=13, and 13 is not divisible by 3.