A number between 300 and 400 is forty 
times the sum of its digits. The tens 
digit is 6 more than the unit digit.
find the number. Found the answer as 360,
but can't figure out all of the steps

Yes a little trial and error gives the
answer.  But we don't want to use any
trial and error.  I also, for personal
reasons. do not want to use anything given
that is not necessary, such as that it is
between 300 and 400, which the other tutor 
used immediately to show that the hundreds
digit is 3.  So I won't even assume that.

It's not easy, though.  Here goes:

Let h = hundredths digit
Let t = tens digit
Let u = units (ones) digit

The first sentence tells us

100h + 10t + u = 40(h + t + u)

100h + 10t + u = 40h + 40t + 40u

60h = 30t + 39u

which can be divided thru by 3

20h = 10t + 13u

The second sentence tells us

t = u + 6

Substituting in 20h = 10t + 13u

20h = 10(u + 6) + 13u

20h = 10u + 60 + 13u

20h = 23u + 60

Write 23u in terms of its nearest 
multiple of 20, which is 20u + 3u

20h = 20u + 3u + 60

Divide thru by 20

h = u + 3u/20 + 3

h - u - 3 = 3u/20

Since the left side is an integer, 
so is the right side

Suppose that integer is A, then

3u/20 = A

3u = 20A

Write 20A in terms of its nearest multiple 
of 3, which is 21A - A

3u = 21A - A

Divide thru by 3

u = 7A - A/3

A/3 = 7A - u

The right side is an integer, so the 
left side must be also.

Suppose that integer is B

A/3 = B

A = 3B

Substitute that in 3u = 20A

3u = 20(3B)

3u = 60B

u = 20B

Now since 0 < u < 9

0 < 20B < 9

Divide thru by 20

0 < B < 9/20

0 < B < 0.45

Since B is an integer, 
it can only be 0

B = 0

substitute in u = 20B

u = 20(0)

u = 0

Substitute in  h = u + 3u/20 + 3

h = 0 + 3(0)/20 + 3

h = 3

Sunstitute u = 0 also in t = u + 6

t = 0 + 6

t = 6

So h = 3, t = 6, and u = 0

So the number is 360.

Notice that we did not need to be given
that the number was between 300 and 400. 

Edwin