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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?

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

Jun 17, 2021
#2
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I tried that and it didn't work.

Guest Jun 17, 2021
#3
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Also, this is modular arithmetic so I think there is a different way which has a different solution.

Guest Jun 17, 2021
#4
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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

Jun 18, 2021
edited by Guest  Jun 18, 2021
edited by Guest  Jun 18, 2021
#7
+35290
+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
+121084
+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

Jun 18, 2021
edited by CPhill  Jun 18, 2021