Let n be the answer.
if you make 78 strings , it is exact.
So n is a multiple of 78, and therefore
n = 78p, for some natural number q
If you make 77 strings of them, you are 50 coins short
So if 50 is added to n, the result is a multiple of 77
Therefore
n + 50 = 77q for some integer
Substitute 78p for n
(1) 78p + 50 = 77q
The smaller coefficient of a letter in absolute value is 77
So we rewrite 78 and 50 in terms of their nearest multiples of 77
We rewrite 78 as 77+1 and 50 as 77-27
(77+1)p + 77-27 = 77q
77p + p + 77 - 27 = 77q
Divide through by 77
p + + 1 - = q
Isolate the fractional terms
- = q - p - 1
The right side is an integer. Let that integer be A.
(2) q - p - 1 = A
and so is the left side
- = A
Clear of fractions
p - 27 = 77A
p = 77A + 27
Substitute in (2)
q - p - 1 = A
q - (77A + 27) - 1 = A
q - 77A - 27 - 1 = A
q - 77A - 28 = A
q = 28 + 77A
So the solution to (1) is (p,q) = (77A+27,28+77A)
n = 78p = 78(77A+27) = 6006A+2106
The smallest solution is when A = 0,
n = 2106.
But there are infinitely many solutions, as A can be chosen as
any positive integer.
Edwin