Maths question(New operations)

x*0 = 1 ... (1)
(x*y)*z = z*(xy) + z ... (2)
Put y = 0, (x*0)*z = z*(0) + z
1*z = 1 + z ...(3)
Put x = 1, y = Y - 1, z = 1 in (2)
[1*(Y - 1)]*1 = 1*(Y - 1) + 1
(1 + Y - 1)*1 = 1 + Y - 1 + 1; both sides use (3)
Y*1 = Y + 1 ... (4)
Put x = 9, y = 10, z = 1 in (2),
(9*10)*1 = 1*(90) + 1
(9*10) + 1 = 91 + 1; LHS uses (4) and right side uses (3)
9*10 = 91

(1)x*0=1 for any values of x;
(2)(x*y)*z=z*(xy)+z

首先，對任意 z,
1*z = (x*0)*z = z*(0) + z = 1 + z

所以，對所有 k ≥ 1,
k*(k+1) = (1*(k-1))*(k+1)
= (k+1)*(k-1) + (k+1)
= (1*k)*(k-1) + (k+1)
= (k-1)*k + 2k

因此，n*(n+1)
= 0*1 + 2{1+ ...+n}
= n^2 + n + 1

取 n = 9, 即得 9*10 = 91.