420
There are only two prime numbers that are consecutive numbers, 2 and 3. Their product is 2 x 3 = 6. The first prime numbers are 2, 3, 5, and 7 and the only two consecutive prime numbers whose product is a single digit are 2 and 3. (The next two consecutive prime numbers, 3 and 5, have a two-digit product.)
There is no set of 3 consecutive odd numbers whose product is 6,873. However, there is a set of 3 consecutive odd numbers whose sum is 6,873: 2289, 2291, and 2293.
There are no such numbers. The smallest set of three consecutive composite numbers is {8, 9, 10} and the product of these numbers is 720.
Let the three integers be, n, (n + 1), and (n + 2) Then at least one of these numbers is even and therefore has a factor of 2. And one of the numbers is divisible by 3 **. Therefore the product has factors of 2 and 3 and is thus divisible by 2 x 3 = 6. ** Either n is divisible by 3. Or, n leaves a remainder of 1 when divided by 3 in which case (n + 2) is divisible by 3. Or, n leaves a remainder of 2 when divided by 3 in which case (n + 1) is divisible by 3.
18
No. Every third consecutive natural number is divisible by 3.
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.
420
No.
There are only two prime numbers that are consecutive numbers, 2 and 3. Their product is 2 x 3 = 6. The first prime numbers are 2, 3, 5, and 7 and the only two consecutive prime numbers whose product is a single digit are 2 and 3. (The next two consecutive prime numbers, 3 and 5, have a two-digit product.)
There is no set of 3 consecutive odd numbers whose product is 6,873. However, there is a set of 3 consecutive odd numbers whose sum is 6,873: 2289, 2291, and 2293.
There are no such numbers. The smallest set of three consecutive composite numbers is {8, 9, 10} and the product of these numbers is 720.
Let the three integers be, n, (n + 1), and (n + 2) Then at least one of these numbers is even and therefore has a factor of 2. And one of the numbers is divisible by 3 **. Therefore the product has factors of 2 and 3 and is thus divisible by 2 x 3 = 6. ** Either n is divisible by 3. Or, n leaves a remainder of 1 when divided by 3 in which case (n + 2) is divisible by 3. Or, n leaves a remainder of 2 when divided by 3 in which case (n + 1) is divisible by 3.
3 x 4 x 5 = 60 The three consecutive numbers are 3, 4, and 5.
5+2+1=8 and 8 is not divisible by 3.
17,19,21