a brick has a length of 2 2/5 cm and a width of 4/5 cm and a height of 1 cm. how many 1/5 cm cubes can fit along the length of t
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Naturally, any integer 
larger than 127 will return 
, and of course 
, so we restrict the possible solutions to 
.
Now,
is the same as saying there exists some integer
such that
We have
which means that any that satisfies the modular equivalence must be a divisor of 120, of which there are 16: 
.
In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,
(If we're allowing
, then I see no reason we shouldn't also allow 2, 3, 4, 5, 6.)
Now,
is the same as saying there exists some integer
We have
which means that any
In the cases where the modulus is smaller than the remainder 7, we can see that the equivalence still holds. For instance,
(If we're allowing
