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
7*8=56
There are no two consecutive primes whose sum, difference, product or quotient is 56.However, there may be some non-standard binary operation such that two primes can be combined to make 56.
7 & 8
It is: (77+32)*(56-45) = 1199
27 29
The two numbers with the sum of 15 and the product of 56 are: 7 + 8 = 15 7 x 8 = 56
X + Y = 56 [therefore] X = 56 - Y XY = 783 [therefore] (56 - Y)(Y) = 783 56Y - Y^2 = 783 solve for Y, then don't forget to plug Y back into the original and find "X" (X = 56 - Y ) a+b = 56 AND: ab = 783 substitution: a = 56-b (from addition equation) put this back into the multiplication equation: (56-b)b = 783 expansion: 56b -b^2 = 783 rearrange: -b^2 + 56b - 783 quadratic formula: b = (-56+sqrt(3136-4(-1)(-783))) / -2 and b = (-56-sqrt(3136-4(-1)(-783))) / -2 b=27 and b=29 from the addition equation: 56-27 = a=29 56-29 = a =27 it's kind of a "repeat" question
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.
7*8=56
There are no two consecutive primes whose sum, difference, product or quotient is 56.However, there may be some non-standard binary operation such that two primes can be combined to make 56.
-56 and -1 -56+(-1)=-57 -56*(-1)=56
7 & 8
TOTAL IS 150
7 and 8
They are: 7 and 8