514 is divisible by these numbers: 1 2 257 and 514.
Between 1 and 2018, 1008 are not divisible by 2.1613 are not divisible by 5.
To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3→ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3→ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3→ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3→ 121 is not divisible by either 2 or 3, so it is not divisible by 6
It's not. If n = 2, then 3n - 1 = 3*2 - 1 = 6 - 1 = 5, which isn't divisible by 2.
338 is divisible by 1, 2 and 338 itself.
4 is divisible by 1, 2 and 4.
No odd numbers are evenly divisible by 2. 3 ÷ 2 = 1 1/2 11 ÷ 2 = 5 1/2
between 1 and 600 inclusive there are:300 numbers divisible by 2200 numbers divisible by 3100 numbers divisible by both 2 and 3400 numbers divisible by 2 or 3.
4 is divisible by 1, 2 and 4 because 1 x 4 and 2 x 2 = 4
514 is divisible by these numbers: 1 2 257 and 514.
Let three consecutive integers be n, n+1 and n+2. If n is divisible by 3 then n+1 and n+2 cannot be divisible by 3 as these numbers will respectively leave remainders of 1 and 2. If n is not divisible by 3 then it will leave a remainder of 1 or 2. If n leaves a remainder of 1, then n+1 leaves a remainder of 2 and n+2 is therefore divisible by 3. If n leaves a remainder of 2, then n+1 is divisible by 3 and n+2 is not divisible by 3 as it leaves a remainder of 1.
There is something missing from your question: what about the two consecutive integers? Is the proof required that one of them is divisible by 2? Or that their product is (which amounts to the same thing)? Showing that one of two consecutive number is divisible by 2: Suppose your two numbers are n and (n+1). If n is divisible by 2, ie n = 2k, the result is shown. Otherwise assume n is not divisible by 2. In this case n = 2m+1. Then: (n+1) = ((2m+1)+1) = 2m + 2 = 2(m+1) which is a multiple of 2 and so divisible by 2. QED. Showing that exactly one of two consecutive integers is divisible by two is shown above with the addition to the first part: "as (n+1) = 2k+1 is not divisible by two and so only n is divisible by 2." To show the product is divisible by 2, show either n is divisible by 2 or (n+1) is as above, then the result follows as one of n and (n+1) is divisible by 2 and so their product is.
Between 1 and 2018, 1008 are not divisible by 2.1613 are not divisible by 5.
To be divisible by 6, the number must be divisible by both 2 and 3:To be divisible by 2 the last digit must be even, ie one of {0, 2, 4, 6, 8};To be divisible by 3, sum the digits of the number and if this sum is divisible by 3, then the original number is divisible by 3.As the test can be repeated on the sum, repeat the summing until a single digit remains; only if this number is one of {3, 6, 9} is the original number divisible by 3.If the number is not divisible by 2 or 3 (or both) then the number is not divisible by 6.examples:126Last digit is even so it is divisible by 2 1 + 2 + 6 = 9 which is divisible by 3, so it is divisible by 3→ 126 is divisible by both 2 and 3, so it is divisible by 6124Last digit is even so it is divisible by 2 1 + 2 + 4 = 7 which is not divisible by 3, so it is not divisible by 3→ 126 is divisible by 2 but not divisible by 3, so it is not divisible by 6123Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 3 = 6 which is divisible by 3, so it is divisible by 3→ 123 is divisible by 3 but not divisible by 2, so it is not divisible by 6121Last digit is not even so it is not divisible by 2 We can stop at this point as regardless of whether it is divisible by 3 or not, it will not be divisible by 6. However, for completeness:1 + 2 + 1 = 4 which is not divisible by 3, so it is not divisible by 3→ 121 is not divisible by either 2 or 3, so it is not divisible by 6
It's not. If n = 2, then 3n - 1 = 3*2 - 1 = 6 - 1 = 5, which isn't divisible by 2.
A number is divisible by 6 if it is divisible both by 2 and 3 1240 is even then is divisible by 2 1+2+4+0 = 7 which is not divisible by 3 then 1240 is not divisible by 3 Thus 1240 is not divisible by 6
338 is divisible by 1, 2 and 338 itself.