(1)
aは素数ゆえ、N = 2(1とa自身)
<a> = 40 = 80/2 = (1 + 79) / 2
∴a = 79
(2)
bは素数ゆえ、b^2の約数は1, b, b^2
∴N = 3
<b^2> = 19 = 57/3
b^2 + b + 1 = 57, b^2 + b - 56 = 0, (b + 8)(b - 7) = 0
bは素数ゆえ、b = 7
(3)
c, d(c < d)は異なる素数ゆえ、cdの約数は1, c, d, cd
∴N = 4
<cd> = 12 = 48/4
cd + c + d + 1 = 48
c(d + 1) + (d + 1) = 48
(c + 1)(d + 1) = 48
c < dより
∴(c + 1, d + 1) = (1, 48), (2, 24), (3, 16), (4, 12), (6, 8)
(c + 1, d + 1) = (1, 48)はc = 0より不適
(c + 1, d + 1) = (2, 24)はc = 1より不適
(c + 1, d + 1) = (3, 16)はd = 15より不適
(c + 1, d + 1) = (4, 12)はc = 3, d = 11より適する
(c + 1, d + 1) = (6, 8)はc = 5, d = 7より適する
∴(c, d) = (3, 11), (5, 7)
(c, d) = (3, 11)のとき、<cd> = (1 + 3 + 11 + 33)/4 = 12
(c, d) = (5, 7)のとき、<cd> = (1 + 5 + 7 + 35)/4 = 12
お礼
ありがとうございます ものすごくわかりやすいです!!!!