A wealthy merchant arranged a large caravan of camels loaded with various goods?
The goods vary from sacks of cheap hay to bags of valuable jewelry, so for convenience of accounting the merchant loaded
1st camel with 10dinar worth of goods,
2nd camel with 20dinar worth of goods,
3rd camel with 30dinar worth of goods,
and so on, each next camel carries 10 dinar worth of goods more.
http://www.ngsprints.co.uk/images/M/756029.jpg
The merchant must pay duty 1 dinar per camel, so at the city gates he declared that he had 450 camels and paid 450 dinars tax, and (plus) gave the tax collector 50 dinars bribe.
The tax collector was very pleased with the bribe, and asked the merchant:
“How do you manage to oversee such huge amount of goods during the jorney?”
The merchant replied honestly:
“Simple. I ride on one those camels, chosen so that the same amount worth of goods is in front of me, as amount worth of goods behind me”
How much money did the merchant save by bribing the the tax collector?
so the merchant has more than 500 camels, right?
let’s say he rode the nth camel.
front are 1st till (n-1)th, and back are (n+1)th till pth, p > 500.
(n-1)[1+(n-1)]/2 = (p-n)[(n+1)+p]/2
n(n – 1) = (p – n)(p + n) + p – n = p^2 – n^2 + p – n
2n^2 = p(p + 1)
(p + 1/2)^2 = 2n^2 + 1/4
(2p + 1)^2 = 8n^2 + 1
letting P = 2p + 1,
P^2 = 8n^2 + 1
{
x^2 = 1 + 8y^2 are
(3,1), (17,6) and (99,35).
}
the above are from Dr. D’s answer.
but i got (n,P) = (204,577) too.
and then some more.
(n,P) = (1189,3363) and (6930,19601)
formula is n(k+1) = 2*[P(k)] + n(k-1).
OR P(k+1) = 6*[P(k)] – P(k-1)
for p > 500, P = 2p+1 > 1001.
we get the minimum P = 3363.
p = (P-1)/2 = 1681
the merchant had 1681 camels in all.
he saved = 1681 – 450 – 50 = 1181 dinars.
answer = 1181 Dinars, riding the 1189th camel.
45000 Kuwaiti Dinar car!!!!
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