You have ten stacks of identical looking gold coins.
Nine of the stacks contain all real gold coins, and one of the stacks is made up entirely of fake gold coins. Each real gold coin weighs exactly 1 gram, while each fake gold coin weighs exactly 1.1 grams. Using a scale, how can you determine the stack of fake coins with only one measurment?

Asslamo 3laykom

dear sattia you are a killer when you answer and when you ask :D

any way i have the answer if the number of the coins in each stuck at least is 10 coins

ok , here it is we will give each stuck number from 1 to 10 , from stuck number 1 we will take one coin

from stuck number 2 we will take 2 coins

from stuck number 3 we will take 3 coins

from stuck number 4 we will take 4 coins
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from stuck number 10 we will take 10 coins

here it is now , if all of the stucks are identical the measurment must give us
(10+9+8+.....+1 ) = 55 grams , but there is a killer stuck from killer riddle from a killer man that must rise the weight so , if the measurment was ,

- 56 grams that means the stuck number 1 is the fake gold stuck

- 57 grams that means the stuck number 2 is the fake gold stuck

- 58 grams that means the stuck number 3 is the fake gold stuck
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- 65 grams that means the stuck number 10 is the fake gold stuck ;)


peace be upon you

Im afraid that it is an incorrect answer because
the maximum weight u can get by this method is 1 + 2 + ... + 9 + 11 = 56 gms
so try again

this maximum happens when the following series is a maximum too:
1 * 1gm + 2*1gm + ... + n*1.1gm + ...
this sum is a max when n is a max and in our riddle n is 10 so the max sum is
1 + 2 + ... + 9 + 10 * 1.1 (11) = 56 gms

Salamo 3alaikom,

Try to make a translation of the riddle in arabic.

Salam.

يوجد لديك عشرة أكوام من العملات الذهبية المتشابهة
تسعة من هذه الأكوام ت�_توى علي عملات ذهبية �_قيقية
ووا�_دة من هذه الأكوام مكونة من عملات ذهبية مزيفة
العملة الذهبية ال�_قيقية تزن وا�_د جرام بينما العملة المزيفة تزن وا�_دووا�_د من عشرة جرام
باستخدام الميزان مرة وا�_دة فقط كيف تستطيع اكتشاف الذهب المزيف؟

why are you trying to make it look as it is complicated

the answer is right with a small calculation mistake (1+1+1 .... instead of 0.1+0.1+0.1 ....) so you can not give the guy zero for his answer

let the answer be:
if the weight is 55.1 g then its stack no. 1
55.2 g its stack no. 2
55.3 g stack no.3 and so on till stack no. 10 with 56 grams

Asslamo 3laykom

ok , i always do it and fall in arithmetical mistake , any way with the previous method of answer here is the answer whith another words

- the weight of the coins we collect from the stucks if all of them are identical wil be :

[(1*1g)+(2*1g)+(3*1g)+(4*1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1g)]=55grams

now here is our cases ,


if the fake coins was in stuck number one the weight of the stuck will be

- [(1*1.1g)+(2*1g)+(3*1g)+(4*1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1g)]=55.1grams


if the fake coins was in stuck number two the weight of the stuck will be

- [(1*1g)+(2*1.1g)+(3*1g)+(4*1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1g)]=55.2grams


if the fake coins was in stuck number three the weight of the stuck will be

- [(1*1g)+(2*1g)+(3*1.1g)+(4*1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1g)]=55.3grams


if the fake coins was in stuck number four the weight of the stuck will be

- [(1*1g)+(2*1g)+(3*1g)+(4*1.1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1g)]=55.4grams

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if the fake coins was in stuck number ten the weight of the stuck will be

- [(1*1g)+(2*1g)+(3*1g)+(4*1g)+(5*1g)+(6*1g)+(7*1g)+(8*1g)+(9*1g)+(10*1.1g)]=56grams


peace be upon you

I didnt
he just solved half the riddle
his explanation was not correct
so it is not solved

Asslamo 3laykom

Originally posted by sattia
يوجد لديك عشرة أكوام من العملات الذهبية المتشابهة
تسعة من هذه الأكوام ت�_توى علي عملات ذهبية �_قيقية
ووا�_دة من هذه الأكوام مكونة من عملات ذهبية مزيفة
العملة الذهبية ال�_قيقية تزن وا�_د جرام بينما العملة المزيفة تزن وا�_دووا�_د من عشرة جرام
باستخدام الميزان مرة وا�_دة فقط كيف تستطيع اكتشاف الذهب المزيف؟

sattia in which language do you talk :D

peace be upon you

utf8