The sum of two numbers is 27. If the biggest divides with the smaller one the quotient becomes 3 and the remainder 3. What are those numbers?

#color(blue)("Setting up the intial conditions")#

Note: the remainder can also be divided into appropriate parts.

Let the lesser value be #a#
Let the greater value be #b#

#color(purple)("Remainder divided into "b" parts")#
#a/b=3+color(purple)(obrace(3/b))#

#a/b=(3b)/b+3/b#

#a=3b+3" ".........Equation(1)#

#a+b=27" "..............Equation(2)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Solving for "a and b)#

Consider #Eqn(2)#

#a+b=27 color(white)("d") -> color(white)("d") a=27-b" "....Equation(2_a)#

Using #Eqn(2_a)# substitute for #a# in #Eqn(1)#

#color(green)(color(red)(a)=3b+3 color(white)("dddd") ->color(white)("dddd")color(red)(27-b)=3b+3 )#

#color(white)("ddddddddddd.d") ->color(white)("dddd") 4b=24#

#color(white)("ddddddddddd.d") ->color(white)("dddd") b=24/4 = 6#

Thus #a=27-6 = 21#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#

Given #a+b=27#

#"Left hand side "6+21->27# so #LHS=RHS#

Given #a/b=3 " remainder "3#

#21-:6 = 3 " remainder "3# sh #LHS=RHS#

The easiest way to solve this problem is by using logic.

If it weren't for that remainder of #3#, the two numbers would be divisible evenly by #3#.

The larger number would be exactly #3# times the smaller number if it weren't for that remainder.

So forgetting about that remainder for a minute, the pair of numbers would be one of the pairs on this list -- numbers exactly divisible by #3#:

3/1=3

6/2= 3

9/3 = 3

12/4 = 3

15/5 = 3

18/6 = 3 #larr# This is the right division not counting the remainder

21/7 = 3

24/8 = 3

and so on.

Search the list to find which pair adds up to exactly #24#.
This works because when you add back the remainder of #3#, they will add up to #24 + 3 =27# as specified in the problem.

You can see right away that #18 + 6=24#

So if you add the remainder of #3# back in, the numbers become #21 + 6= 27#

#(18+3) -:6 = 3  "remainder"  3#

This answer satisfies both of the requirements of the problem.

1) The quotient of #21-:6# is #3 "remainder" 3# as the problem specifies.

2) The sum of #21+6= 27#, as the problem specifies

Answer
The two numbers are #21# and #6#

#color(white)(mmmmmmmm)#―――――――――

The answer you reached by using logic can be used to find the way to write the equation. Writing the equation is the hard part, and it might be the only solution method the professor will accept.

Let #x# represent the divisor. That makes the dividend #3x + 3.#

#(3x + 3)##larr# dividend
#color(white)()#――――
#color(white)(llll)##(x)# #larr# divisor

This division will give a quotient of #3# with #3# as a remainder.

The problem also specifies that these two amounts add up to #27#
#(3x + 3) + (x) = 27#

Solve for #x#, already defined as the smaller number.

This works out to
#x = 6#, which means that #(3x + 3)# (the larger number) must be #21#

Same answer
The two numbers are #21# and #6#