You have seven bags of gold coins. Each bag has the same number of gold coins. One day, you find a bag of 53 coins. You decide to redistribute the number of coins you have so that all eight bags you hold have the same number of coins. You successfully manage to redistribute all the coins, and you also note that you have more than 200 coins. What is the smallest number of coins you could have had before finding the bag of 53 coins?

P.S The answer by EP here: https://web2.0calc.com/questions/need-help-fast-please_1 is incorrect.

Guest Jun 17, 2021

#1**+1 **

I believe my answer is correct 29 coins per bag 7 bags x 29 coins/bag = 203 coins to start .....if you complete the math (which you should be able to do)

IF you have 200 coins after adding the 53 coins then the answer would be 147 total coins....but the question states GREATER than 200 coins.

~EP

Guest Jun 17, 2021

#3**0 **

Also, this is modular arithmetic so I think there is a different way which has a different solution.

Guest Jun 17, 2021

#4**+2 **

You tried 203 coins and that didn't 'work' ?? Let's re examine:

x has to be divisible by 7 x + 53 has to be divisible by 8 and x + 53 has to be the least number above 200

208 is the first number greater than 200 that is divisible by 8

but (208 - 53) is not divisible by 7 so it is not 208

(216-53) is not divisible by 7

224-53 is not divisible by 7

232-53 is not divisible by 7

240-53 is not divisible by 7

248-53 is not divisible by 7

256-53 IS divisible by 7 this is the lowest number greater than 200 that is divisible by 8 and 256-53 is divisible by 7

(256-53)/7 = 29 coins per the original bags 29 * 7 = 203 coins originally THERE is no other answer ..... sorry !

As I stated earlier IF YOU HAVE 200 coins at the end (start with 200 -53 = 147 coins) ....that would work too BUT THE QUESTION STATES > 200 coins at the end

Guest Jun 17, 2021

#5**+1 **

8th bag has 53 coins. And in total we have more than 200 coins.

: 7y + 53 ≥ 200.

7y ≥ 200-53 ; 7y ≥ 147

y ≥ \( {147 \over 7}\)

y ≥ 21. : Smallest number of coins before Friday bag y 53 coins is given by :

7 * 21

= 147

Guest Jun 18, 2021

edited by
Guest
Jun 18, 2021

edited by Guest Jun 18, 2021

edited by Guest Jun 18, 2021

#7**+1 **

Sorry....as I noted in my answer above....147 coins originally is incorrect because 147 + 53 = 200 and you have to have MORE THAN 200 coins after adding the 53 .......

ElectricPavlov
Jun 18, 2021

#6**+1 **

Let x be the number of coins in each of the 7 bags originally

So

7x + 53 > 200

7x > 147

x > 21

Since we have 8 bags, then the number of coins must be an even number > 200

Through a little trial and error, I find that the number of coins = 256.....and there were 29 coins in each of the 7 bags originally

Logic :

Let 29 be the number of coins in each of the 7 bags originally

Distribute 3 coins out of the 53 into each of the 7 bags

Then each of the 7 bags contains 32 coins and the 8th bag contains 53 - 7(3) = 53 - 21 = 32 coins

So....as found previously.....we must have had 29 x 7 = 203 coins originally

CPhill Jun 18, 2021