Gold Coin Transporters

Number Theory Level 2

Three gold coins of weight 780 g, 840 g, and 960 g are cut into small pieces of equal weight. If it takes 2 people to transport one piece of gold, what is the fewest number of people that are needed to transport all these pieces?


The answer is 86.

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Aditya Raj by Aditya Raj , 22, India

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2 solutions

Parth Lohomi
Nov 19, 2014

HCF ( 780 , 840 , 960 ) (780, 840, 960) = 60 Thus total number of pieces

780 60 \dfrac{780}{60} + 840 60 \dfrac{840}{60} + 960 60 \dfrac{960}{60} = 13 + 14 + 16 13+14+16 = 43 43

total persons required = 43 ∗ 2 43*2 = 86 \boxed{86}

24 Helpful 5 Interesting 3 Brilliant 1 Confused

For people to be min gold pieces have to be min. And for the gold pieces to be min each piece must carry max weight. So, we take GCD of three weights. Such that a greatest common weight can be found among the three gold pieces.

Abhijeet Kushwaha - 10 months, 3 weeks ago

I didn't get it, because if we need MAX weight, why we would chose the greatest divisor? The more we divide it, the more soft pieces get. 2 is a common divisor that gives weights far greater than 60, and it demands just 12 peoples.

Felipe Merique - 1 month, 2 weeks ago
Anatoliy Razin
Nov 18, 2014

2 ⋅ 780 + 840 + 960 g c d ( 780 , 840.960 ) = 86 2 \cdot \frac{780 + 840 + 960}{gcd(780, 840. 960)} = \boxed{86}

6 Helpful 0 Interesting 0 Brilliant 2 Confused

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