Let p, p+2, and p+4 be three consecutive odd numbers. Then, if p ≡ 0 (mod 3), p is divisible by 3; if p ≡ 1 (mod 3), then p+2 is divisible by 3, and if p ≡ 2 (mod 3), then p+4 is divisible by 3.
That is, at least one of p, p+2, and p+4 must be divisible by 3. The only prime that can be divisible by 3 is 3 itself; for any other positive integer divisible by 3 must have another factor, making it composite. This gives the possibilities
-1, 1, 3
1, 3, 5
3, 5, 7
The first two are eliminated because -1 and 1 are not primes, leaving 3, 5, 7 as the only set of three consecutive integers that are all prime.
504, 756, 1008
44
-13, -15 and -17
91
1422, 1431, 1440
No.
5711
There are not three odd primes with the sum of 14. The sum of three odd primes will be an odd number.
As the number 1 is considered a special case and not a prime, there cannot be three consecutive numbers that are prime. Any three consecutive numbers must include at least one even number. With the exception of the number 2, no even numbers are prime.
EVERY three consecutive numbers add to a multiple of 3: Proof: numbers are n, n + 1 and n + 2. The total is 3n + 3 or 3(n + 1) This means that for any three consecutive numbers, the total is 3 times the middle number.
No. Any three consecutive numbers will have at least one of them which is divisible by 2, which means it cannot be prime. And since 1 is not considered a prime number, it cannot happen.
The sum of three consecutive odd numbers must be divisible by 3. As 59 is not wholly divisible by 3 the question is invalid. PROOF : Let the numbers be n - 2, n and n + 2. Then the sum is 3n which is divisible by 3. If the question refers to three consecutive numbers then a similar proof shows that the sum of these three numbers is also divisible by 3. Again, the question would be invalid.
If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.If you take three consecutive odd (or three consecutive even) numbers, one of the three will always be a multiple of 3.
This question cannot be answered because three odd primes always have an odd number for their sum.
3
There is only one pair of consecutive prime numbers, and the prime numbers are two and three, because any pair of consecutive numbers has one odd and one even number, and two is the only even prime number, because all other even numbers can be divided by two, and the only pairs of consecutive numbers are one and two and three, but one is not prime because it only has one factor, thus making the only consecutive pair of primes two and three. But the problem asks for the product of the two numbers, not the numbers themselves, so just multiply two and three together to get a final result of six.
There is no set of three consecutive integers for 187.