Organisers for a polytechnic graduation has three sections of chairs to seat graduates, faculty and invited. 1120 chairs have been arranged for the graduatesm 1400 for the invited guests and 896 for the lecturers, if all the rows have the same number of chairs, calculate the greatest number of chairs in each row if no chairs are left over?
1120 = 56*20
1400 = 56*25
896 = 56*16
No number bigger than 56 divides all three, so 56 is the greatest number of seats per row.
Sir steve help me answer my post please.
Man, I've been trying. I just can't seem to get a handle on how the distance from the vertex to the incenter can be used to get the third side. If altitudes (of distance r) are dropped from the incenter to the sides, and we label the sides of the kite-shaped pieces as x,y,z, we have
2(x+y+z) = 39
x^2+r^2 = 16
y^w+r^2 = 81
But we only have three equations, and I can't relate the inradius r to the given.
Got me stumped so far.
To calculate the greatest number of chairs in each row without any chairs left over, we need to find the greatest common divisor of the three numbers: 1120, 1400, and 896.
One way to find the greatest common divisor (GCD) is by using prime factorization.
First, let's factorize each of the three numbers:
1120 = 2 * 2 * 2 * 2 * 5 * 7
1400 = 2 * 2 * 2 * 5 * 5 * 7
896 = 2 * 2 * 2 * 2 * 2 * 7
Now we can find the common factors among these factorizations.
The common factors are 2 * 2 * 2 * 7, which is equal to 56.
Therefore, the greatest number of chairs in each row without any chairs left over is 56.