Question

abc is a three digit number. ab, bc, ca are two digit numbers. Determine all three digit numbers abc such that abc = ab + bc + ca​

Answer 1

Answer:

Let ∆ = | a a3 a4 - 1 b b3 b4 - 1 c c3 c4 - 1 |

then,∆ = | a a3 a4 - 1 b b3 b4 - 1 c c3 c4 - 1 | + | a a3 -1 b b3 -1 c c3 -1 |

⇒ ∆ = abc | 1 a2 a3 1 b2 b3 1 c2 c3 | - | a a3 1 b b3 1 c c3 1 |

Applying R2 = R2 - R1 & R3 = R3 - R1

⇒∆ = abc | 1 a2 a3 1 b2- a2 b3 - a3 1 c2 - a2 c3 - a3 | - | a a3 1 b - c b3 - a3 0 c - a c3 - a3 0 |

⇒∆ = abc (b - a) (c - a) | 1 a2 a3 0 b + a b2 + a2 + ab 0 c + a c2 + a2 + ac | - (b - a) (c - a) | a a3 1 1 b2 + a2 + ab 0 1 c2 + a2 + ac 0 |

⇒ ∆ = abc (b - a) (c - a) | b+a a2+b2+ab c+a a2+c2+ac | -(b - a) (c - a) | 1 a2+b2+ab 1 a2+c2+ac |

Apply R2 = R2 - R1

⇒∆ =abc (b - a) (c - a) | a+b a2+b2+ab c-b c2-b2+a(c-b) | - (b - a) (c - a)| 1 a2+b2+ab 0 c2-b2+a(c-b) |

⇒∆ = abc (b - a) (c - a)(c-b)| a+b a2+b2+ab 1 a+b+c | - (b - a) (c - a)(c-b)| 1 (b - a) (c - a) 0 a+b+c |

⇒∆ = abc (b - a) (c - a)(c-b)[(a+b)(a+b+c)-(a2+b2+ab)] - (b - a) (c - a)(c-b)(a+b+c)

⇒∆ = abc(a-b)(b-c)(c-a)[ab+bc+ca] - (a-b)(b-c)(c-a)(a+b+c)

⇒∆ =(a-b)(b-c)(c-a)[abc(ab+bc+ca)-(a+b+c)] Now, ∆ = 0

⇒(a-b)(b-c)(c-a)[abc(ab+bc+ca)-(a+b+c)] = 0

⇒[abc(ab+bc+ca)-(a+b+c)] = 0 [because a≠b; b≠c; c≠a, therefore a-b≠0; b-c≠0; c-a≠0]

⇒abc(ab+bc+ca) = a+b+c

⇒ab+bc+ca=a+b+c/abc

hupe this helps you a bit as i can't complete this due to wifi problems and the app is crashing a lot

sorry fir trouble but d continue usinf thjs method tou will get yojr answer soon☺☺☺☺☺☺☺☺☺