A band of 17 pirates stole a sack of gold coins. When they tried to divide the fortune into equal portions, 3 coins remained. In the ensuing brawl over who should get the extra coins, one pirate was killed. The wealth was redistributed, but this time an equal division left 10 coins. Again an argument developed in which another pirate was killed. But now the total fortune was evenly distributed among the survivors. What was the least number of coins that could have been stolen

The least number of coins that could have been stolen is 3,930.

Step-by-step explanation:

In the beginning, there are N coins to be distributed among 17 pirates, and we know that if we divide evenly these coins, there are 3 coins remaining.

Then we can write the total number of coins as a multiple of 17 plus 3.

N = 17*k + 3

where k is an integer.

After that, a pirate is killed, so now we have 16 pirates, and now there are 10 coins left, so now we can write N as a multiple of 16 plus 10:

N = 16*j + 10

where j is an integer.

Finally, having 15 pirates, the total number of coins can be divided evenly, then N is a multiple of 15

N = 15*m

where m is an integer.

So we have these 3 equations:

N = 17*k + 3

N = 16*j + 10

N = 15*m

So we can express equations like:

15*m = 16*j + 10

15*m = 17*k + 3

Where we need to find solutions such that both of these variables are integers.

For the first one we can write:

15*m - 16*j = 10

One way to do it, is to write this like a linear equation:

m = (16*j + 10)/15

Graph this, and find the pairs of points in the line that have bot integer values, or we can just find some values of j such that:

16*j + 10 is a multiple of 15.

For example, with j = 5 we have:

m = (16*5 + 10)/15 = 6

so j = 5 and m = 6 is a possible solution.

Now if we use m = 6 in the other equation:

15*m = 17*k + 3

we can see if k is also an integer:

k = (15*m - 3)/17

replacing m by 6:

k = (15*6 - 3)/17 = 5.1

This solution does not work.

Let's find others:

if j = 20  (at this point we already know that the possible values of j "jump" by 15 units, like 5 + 15 = 20) then:

m = (16*20 + 10)/15 = 22

Then:

j = 20 and  m = 22 is another solution.

Same as before, let's use m = 22 in the equation for k:

k = (15*22 - 3)/17 = 19.2

Let's find another solution for m.

if  j = 35

m = (16*35 + 10)/15 = 38

Using this value of m in the k equation we get:

k = (15*35 - 3)/17 = 33.4

Let's find another solution for m.

if we take j = 50 we get:

m = (15*50 + 10)/15 = 54

Using this in the k equation we get:

k = (15*54 - 3)/17 = 47.5

Eventually, for j = 5  + 15*16 = 245 we get:  (arrived by iteration, just try each time using the previous value of j plus 15)

m = (16*245 + 10)/15 = 262

replacing this in the k equation we can find:

k = (15*262 - 3)/17 = 231.

So the first solution is:

j = 245, m = 262, k = 231

Then:

N = 262*15 = 3,930

The least number of coins that could have been stolen is 3,930.

b

step-by-step explanation:

if a was wrong then it has to be b it cant be c either because not all the y intercepts are the same!

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