5*6*7 = 210
The numbers are 69, 70, and 71.
69, 70, 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
(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
(3/5*210)-(2/3*96) = 62
They are 14 and 15.
The numbers are 14 and 15.
The numbers are 69, 70, and 71.
The numbers are 69, 70 and 71.
702 = 26 x 27; 210 = 14 x 15.
69, 70, 71
Consecutive numbers can't both be multiples of 7. The LCM of consecutive numbers is their product. 14 and 15 are consecutive numbers whose LCM is a multiple of 7 that is greater than 200.
14 x 15 = 210
210 and 211
Three numbers that can be multiplied together to get 210 are 5, 6, and 7. When you multiply these numbers together (5 × 6 × 7), the product is indeed 210. Other combinations, such as 2, 3, and 35, or 1, 10, and 21, can also yield the same result.
To find three numbers that add to 630 and can be represented using multiplication, consider the numbers 210, 210, and 210. While these three numbers add up to 630 (210 + 210 + 210 = 630), their product is (210 \times 210 \times 210). However, if you're looking for a multiplication factor, you could consider (1 \times 1 \times 630) or any variation thereof, where two of the numbers are 1 and the third is 630.
The numbers are 51, 52, 53 and 54.