69, 70, 71
Wiki User
∙ 2008-12-05 00:31:325*6*7 = 210
You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210
The numbers are 69, 70, and 71.
(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
210 = 2*105 = 2*3*35 = 2*3*5*7. To simplify sqrt(210) you need a pair of factors to form a square. You can then take out the square root, which will be a whole number. Since 210 has no pair of factors, sqrt(210) cannot be simplified.
5*6*7 = 210
The numbers are 14 and 15.
They are 14 and 15.
702 = 26 x 27; 210 = 14 x 15.
The numbers are 69, 70 and 71.
You can solve this in two ways.1) Trial and error. That is, try multiplying two consecutive integers; if the product is too large, try smaller integers; if the product is too small, try larger consecutive integers. 2) Call the two consecutive integers "n" and "n+1", and solve the equation: n(n+1)=210
The numbers are 69, 70, and 71.
14 x 15 = 210
Consecutive numbers wouldn't both be multiples of 7. If you mean consecutive multiples of 7, the LCM of 210 and 217 is 6510 which is their product divided by 7.
210 and 211
(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
The product is 210