Question 252900: When n is divided by 5, the remainder is 4. When n is divided by 4, the remainder is 3. If 0 < n < 100, what is one possible value of n ?
Found 2 solutions by checkley77, Theo:
Answer by checkley77(12844)   (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! When n is divided by 5, the remainder is 4. When n is divided by 4, the remainder is 3. If 0 < n < 100, what is one possible value of n ?
n/5 = x + 4/5
n/4 = y + 3/4
multiply first equation by 5 to get:
n = 5x + 4
multiply second equation by 4 to get:
n = 4y + 3
your 2 equations become:
n = 5x + 4
n = 4y + 3
since both expressions equal to n, then they are equal to each other so you get:
5x + 4 = 4y + 3
solve for x to get:
x = (4y-1)/5
if y = 1, then x = 3/5.
the assumption is that you want a whole number (integer) so this answer is not good.
if the numerator was 15, then you would get 15/5 = 3.
set your numerator equal to 15 and solve for y to get:
4y - 1 = 15
add 1 to both sides to get 4y = 16
divide both sides by 4 to get y = 4
if y = 4, then x = 15/5 = 3 which should satisfy the equation.
you get x = 3 and y = 4.
your two original equations were:
n/5 = x + 4/5
n/4 = y + 3/4
substitute 3 for x and 4 for y to get:
n/5 = 3 + 4/5
n/4 = 4 + 3/4
multiply first equation by 5 and second equation by 4 to get:
n = 5*3 + 4
n = 4*4 + 3
solve for n to get:
n = 19
n = 19
one of your possible values for n = 19.
19/5 = 3 with a remainder of 4.
19/4 = 4 with a remainder of 3.
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