Question 1040980: When Joyce counts the pennies in her bank by fives, she has one left over. When she counts them by threes, there are two left over. What is the least possible number of pennies in the bank?
Found 3 solutions by Edwin McCravy, AnlytcPhil, robertb:
Answer by Edwin McCravy(20060)   (Show Source):
You can put this solution on YOUR website! When Joyce counts the pennies in her bank by fives, she has one left over. So she has 5N+1 pennies, where N is some positive integer When she counts them by threes, there are two left over.So she has 3M+2 pennies, where M is some other positive integer.
So 3M+2 = 5N+1
3M-5M = -1
Write 5 in terms of its nearest multiple of 3, which is 6
3M-(6-1)N = -1
3M-6N+N = -1
Divide through by 3
M-2N+N/3 = -1/3
Get fractions left, whole numbers right:
N/3+1/3 = 2N-M
Right side is a positive integer, so the right side is too.
Let that positive integer by A
N/3+1/3 = A; 2N-M = A
N+1 = 3A;
N = 3A-1;
Substitute in 2N-M = A
2(3A-1)-M = A
6A-2-M = A
5A = M+2
5A-2 = M
So N = 3A-1 and M = 5A-2 What is the least possible number of pennies in the bank?The least possible value of A to make N the
least positive integer is A=1.
So N = 3A-1 = 3(1)-1 = 3-1 = 2 and M = 5A-2 = 5(1)-2 = 5-2 = 3
So she has 5N+1 pennies = 5(2)+1 = 10+1 = 11 pennies
As a check:
She also has 3M+2 pennies = 3(3)+2 = 11 pennies.
And when we divide 11 by 5, we get 1 remainder, and
when we divide 11 by 3, we get 2 remainder.
So 11 pennies is correct.
Edwin
Answer by AnlytcPhil(1807)  (Show Source):
Answer by robertb(5830)  (Show Source):
You can put this solution on YOUR website! Let N = the possible number of pennies of Joyce in the bank.
==> N = 5r + 1 for some non-negative integer r.
==> N = 3s + 2 for some non-negative integer s.
==> 5r + 1 = 3s + 2 ===> 5r - 3s = 1,
which is a linear Diophantine equation.
Now any linear Diophantine equation
ar - bs = c
has an integer solution in r and s if and only if gcd(a,b) divides c, and that all integer solutions are of the form
 and  , where  and  are particular solutions.
gcd(5,3) = 1 divides c = 1 ==> there are general solutions for the DE equation.
A particular solution to this equation is  and  .
==> r = 2 + 3k and s = 3 + 5k,
for k = 0, 1, 2, 3, 4, ...
are the general solutions.
r = 2 and s = 3 (when k = 0) give the lowest possible value for N which is
N = 11,
and therefore 11 is the least possible number of pennies in the bank.
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