The product of four consecutive positive integers is 1 less than $461^2$. What is the least of these four numbers?

The least of the four numbers is 20

Step-by-step explanation:

The given information are;

The product of 4 consecutive numbers = 461² - 1

The required information is the least among the four numbers

Let the numbers be N-2, N-1, N, N + 1

We have;

(N - 2)×(N)×(N + 1)(N - 1) = 461² - 1

(N - 2)×(N)×(N + 1)(N - 1) - (461² - 1) = 0

Which gives;

N⁴ - 2·N³ - N² + 2·N - 212520 = 0

Factorizing online also gives using ;

(N + 21)(N - 22)(N² - N + 460) = 0

Therefore;

N = 22 or -21 are possible solutions

However, the requirement for positive integers give the possible solution as N = 22

Therefore, where the four numbers are N-2, N-1, N, N + 1, we then have;

22 -2, 22 -1, 22, 22 + 1 which gives the numbers as 20, 21. 22, and 23

The least of the four numbers is therefore 20.

hi,

6x times 4 is 24x and 5 times 4 is 20, so the answer is 24x+20.

hope this and for a little humor i am groot.

c.

step-by-step explanation:

she should of added 9^2 and 12^2. that is because it is a^2+b^2=c^2. c^2 is the hypotenuse and should always be bigger than the other two numbers. which is why she should of added 9 squared and 12 squared together instead of 15 squared.