(x)(x + 1)(x + 2) = 210
(x^2 + x)(x + 2) = 210
x^3 + 2x^2 + x^2 + 2x = 210
x^3 + 3x^2 + 2x = 210
x^3 + 3x^2 + 2x - 210 = 0
Let's try 5, as a factor of 210.
5] 1 3 2 -210
5 40 -210
---- 8 42 0 remainder, so 5 is a root.
Since 5 is a root of the equation, let say that 5 is the first number. So, the second number is 6, and the third number is 7.
Check:
5 x 6 x 7 = 210
10 and 12
69, 70, 71
The average of 33 consecutive whole numbers is 58, what is the smallest of these whole numbers? The answer is 42
96 cannot lie between two consecutive whole numbers.
The numbers are 88 and 89 (88 x 89 = 7,832). This is very easy to find. The products you are looking for are consecutive and so they are very close to each other. Calculate the square root of the number, and it is easy to see that if there are two consecutive whole numbers that will be products of the number, they will be consecutive whole numbers very close to the square root. The square root of 7832 is about 88.5.
The numbers are 30, 31 and 32.
There is no set of three consecutive whole numbers that add up to 154.
No.
There are no such whole numbers. The sum of three consecutive whole numbers must be a multiple of 3; as 68 is not a multiple of 3 (68 = 3 × 22 2/3) it cannot be the sum of three whole numbers.
-3, -2 and -1.
There is no such number.
1 and 2, whose product is 2.
The sum of 3 consecutive whole numbers is always equal to 3 times the middle number in that sequence.
Yes but the on prime numbers are 2,3
The three consecutive whole numbers you are looking for are 1, 2, and 3. The sum of the first two numbers, 1 + 2 = 3.
four
For the product to be zero, one of the numbers must be 0. So the question is to find the maximum sum for fifteen consecutive whole numbers, INCLUDING 0. This is clearly achived by the numbers 0 to 14 (inclusive), whose sum is 105.