345 and 218
Since both 3 and 5 are prime numbers, only numbers that are multiples of its product are the numbers that are divisible by both. 15 is the LCM of 3 and 5 and hence all multiples of 15 are divisible by both 3 and 5
The numbers that are divisible by both 3 and 5 have to be a factor of 15. That leaves 15, 30, 45, 60, 75, and 90. There are 6 numbers less than 100 divisible by 3 and 5
20 of them.
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.
List the numbers: 216, 360 Divide both numbers by 2, since they are both divisible by 2: 108, 180 Divide both numbers by 2 again, since they are both divisible by 2: 54, 90 Divide both numbers by 2 yet again, since they are both divisible by 2: 27, 45 Divide both numbers by 3, since they are both divisible by 3: 9, 15 Divide both numbers by 3 again, since they are both divisible by 3: 3, 5 Since 3 and 5 do not have any common factors greater than 1, we can now multiply all the numbers we divided by: 2 x 2 x 2 x 3 x 3 = 8 x 9 = 72 GCF(216, 360)=72
Numbers that are divisible by both 3 and 5 must be multiples of the least common multiple of 3 and 5, which is 15. Therefore, any number that is divisible by 15 will also be divisible by both 3 and 5. This includes numbers like 15, 30, 45, 60, and so on.
There is no such number. If you have any such number, n, that is divisible by 3 and 5 then n + 15 is larger, and is divisible by both. And you can add another 15 to that number, and then to that, for ever more.
Clearly, both 810 and 910 are divisible by 5. A wee trick regarding numbers that are a multiple of 3, is that the component numbers, when added together, are divisible by three. So, 654 is divisible by 3, as 6+5+4 = 15 which is divisible by 3. In your example, 810 components add up to 9, which is divisible by 3. But the components of 910 add up to 10 which is not so divisible.
8 of the numbers less than 660 are divisible by 5 and 11 but not 3. All numbers divisible by 5 and 11 are multiples of their lcm; lcm(5, 11) = 55 All numbers divisibile by 5, 11 and 3 are multiples of their lcm; lcm(5, 11, 3) = 165 659 ÷ 55 = 11 54/55 → 11 numbers less than 660 are divisible by 5 and 11 659 ÷ 165 = 3 164/165 → 3 numbers less than 660 are divisible by 5, 11 and 3 → of the 11 numbers less than 660 divisible by 5 and 11, 3 are also divisible by 3 → 11 - 3 = 8 numbers less than 660 are divisible by 5 and 11 but not 3,.
A Venn diagram for numbers divisible by both 4 and 5 would have two overlapping circles. One circle would represent numbers divisible by 4, while the other circle would represent numbers divisible by 5. The overlapping region where the two circles intersect would represent numbers divisible by both 4 and 5. This intersection would include numbers that are multiples of both 4 and 5, such as 20, 40, 60, and so on.
Your question is impossible to answer. Any number that is divisible by both 2 and 5 will also be divisible by 10. 30 and 60 are not divisible by 9.
No, look at 5 it is odd and not divisible by 3. false,because if you look at the 5 it is not divisible by 3.