27 and 29
I did this in my head by applying certain rules: the two numbers had to end in 7 and 9 it's the only way I could think other 1 and 3 to get a 3 at the end of the product they both had to be bigger than 20 and smaller than 30 smaller than twenty and the product is less than 400, bigger than 30 and the product is bigger than 900 After applying these two rules, the only possible answer was
27 and 29
27 + 29 = 56, 27 x 29 = 783
0 and the multiples of 783 are divisible by 783.
783 = 3 * 3 * 3 * 29
Any of its factors will divide into 783 evenly with no remainder
783.00 also means 783 So 783 = DCCLXXXIII
27 + 29 = 56, 27 x 29 = 783
27 29
Let the smaller number be n then the other number is (n + 2). Then, n(n + 2) = 783 n2 + 2n = 783......which can be rearranged as : n2 + 2n - 783 = 0 This can be factored : n2 + 2n - 783 = (n + 29)(n - 27) = 0 The solution that concerns us is when n is positive. This occurs when, n - 27 = 0 : n = 27. Then the two consecutive odd numbers are 27 and 29.
0 and the multiples of 783 are divisible by 783.
783 = 3 * 3 * 3 * 29
Any of its factors will divide into 783 evenly with no remainder
783.00 also means 783 So 783 = DCCLXXXIII
261 multiplied by 3 is 783.
391x2
Their product: 783.
Their product.
261