Question 54023This question is from textbook Algebra Structure and Method
: Section title is Puzzle Problems.
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
This question is from textbook Algebra Structure and Method
Found 2 solutions by Nate, AnlytcPhil:
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! Number ~> a b c
a = 3 because 300 < a < 400
3 b c
300 + 10b + c = 40(3 + b + c)
300 + 10b + c = 120 + 40b + 40c
180 = 30b + 39c
and ~>
b = c + 6
plug ~>
180 = 30(c + 6) + 39c
180 = 30c + 180 + 39c
0 = 69c
0 = c
plug ~>
b = c + 6
b = 0 + 6 = 6
Number ~> 360
Answer by AnlytcPhil(1806)  (Show Source):
You can put this solution on YOUR website! 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
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