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Given that a, B and C are prime numbers, and a + B + C = 86, AB + BC + Ca = 971, find the value of ABC
The sum of the three indices is even
One is even
Let a be even
Then a = 2
b+c=84
2b+2c+bc=971
So BC = 971-2 (B + C) = 803
So ABC = 2 * 803 = 1606
A + B + C = 94, AB + BC + AC = 2075, a, B, C equal to several (ABC is prime)
We know that all prime numbers except 2 are odd. If a + B + C = 94, then one must be 2 [because the addition of three odd numbers cannot be even]. Let C = 2, and the original equation system be changed to a + B = 92. ① AB + 2 (a + B) = 2075, that is, AB + 2 * 92 = 2075, ab = 1891. ② from ①, B = 92-a is substituted into ②, and (92-a) * a = 1891a ^ 2-92a + 189
Given that a, B, C are prime numbers and a + B + C = 66, AB + BC + Ca = 971, find the value of a, B, C
The three prime numbers (a, B, c) are 2, 11, 73
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