Open and Shut Case
Corinna watched Eric
struggle with an armful of books and notebooks as he tried to open his locker.
Finally he dumped his books on the floor, took a slip of paper out of his
pocket, and started turning the dial on his locker. As Corinna approached, he
crumpled the paper and jammed it back in his pocket.
"Can
I help?" Corinna asked.
Eric
looked up and said, "I feel so dumb - I can't always remember my locker
combination."
"Well,
what I do," Corinna said, "is try to find some pattern in the numbers.
For example, my locker combination last year was 24-32-40, or eight multiplied
by three, four, and five."
"That's
nice, but my three numbers are all two-digit numbers, and they don't have any
common factors," said Eric.
"Well,
if you show me the numbers," Corinna answered, "I'll try to help you
think of something."
Eric
held out the piece of paper. He and Corinna were quiet for a few moments as they
considered the numbers. Suddenly Eric shouted, "Perfect squares!"
"Yes,"
said Corinna, "these two numbers are perfect squares."
"But,"
said Eric, "they both have prime square roots, and that is why my numbers
don't have any common factors."
"I've
got it!" exclaimed Corinna. "The middle number is the sum of the
square roots of the other two numbers."
Eric
never forgot his combination again!
Can
you find Eric's combination? Can you find more than one set of numbers that
could be Eric's combination?
Solutions
Method 1
There are two possibilities: 25,12,49 or 49,12,25
I used the process of elimination for this problem. I first replaced
the blanks with variables. I got:
a,b,c
I found a formula to solve b: b = _/a + _/c.
I needed to solve a and c to solve b.
Since they both have the same requirements, I figured what either one
could be. I replaced a and c with d.
d is a two digit number.
d is a perfect square, and it has a prime square root.
d's square root must be less than 10, since the square of ten is 100,
a three digit number.
d's square root could be (must be a prime):
2, 3, 5 or 7.
Let us find their squares:
4, 9, 25 and 49.
4 and 9 are ruled out since they are one digit numbers.
d = 25 or 49
Let us replace 'a' with 25 and 'c' with 49:
25,b,49
b = 5+7 = 12
Therefore, his combination could be 25, 12, 49 or 49, 12, 25.
You may say that it could be 25, 10, 25 or 49, 14, 49.
Those are ruled out because 'a' and 'c' cannot have any common factors,
which they do in those two possibilities (5 and 7 respectively).
Method 2
Eric's combination is 25-12-49, or it could be 49-12-25.
FIrst, I made a list of all the double-digit perfect squares and
their square roots.
Perfect squares Square roots
16 4
25 5
36 6
49 7
64 8
81 9
Then I crossed out all the perfect squares that did not have prime
square roots, since the two perfect squares in Eric's combination
have prime square roots.
Perfect squares square roots
25 5
49 7
Since there are only two perfect squares with prime square roots,
then two of the numbers in Eric's combination are 25 and 49. The
square root of 25 is 5, and the square root of 49 is 7, which are
both prime. 5 + 7 = 12, so the middle number in Eric's combination is
12. That means that Eric's combination is either 25-12-49, or
49-12-25.
Method 3
The locker combination is either 25-12-49 or 49-12-25
Explore: The question is asking what could Eric's combination be.
We know the first and last numbers are perfect squares that have prime square roots.
We know prime numbers can be divided by only one and themselves.
We know the middle number is the sum of the square roots of the other numbers.
We also know the locker combination has three numbers.
We know the numbers are all two digits and they don't have any common factors.
Plan: We'll find the two digit perfect squares. Then we'll throw out
the one's who's square roots aren't primes. Then we'll find the sums
of the square roots. Then we will make sure that they don't have any
common factors.
Solve: 16, 25, 36, 49, 64, and 81 are all two digit perfect squares.
16, 36, 64, and 81 have square roots that are not prime numbers.
That leaves 25 and 49 to be the two end numbers. 5 + 7 = 12, 25-12-49 and 49-12-25
are both acceptable answers.
Examine: We went through all the qualifications we wrote in the explore step,
we made sure we answered the correct question and we reread the problem to
make sure our answer was reasonable.
Divided Rectangle
A rectangle is divided
into four rectangles with areas 45, 25, 15, and x. Find x.
Solutions
Method
1
There are at least 4 solutions to this problem. The area of X is
always 27 and it consists of sides .6 and 45, 3 and 9, 6 and 4.5,
12 and 2.25
First, examine rectangle 45, its whole number factors are 1 X 45,
3 X 15, and 5 X 9.
Second, examine rectangles 25 and 15 the same way.
Rectangle
45 25 15
1 1 1
3 5 3
Factors: 5 25 5
9
15
Next, determine the common factors of the rectangles 45 and 25,they
are 1 and 5 using 1 as the height of 1 side, the lengths of the top rectangles
are 45 and 25. Dividing Rect 15 by 25 results in the second side
of height .6. Rectangle X has sides of 45 and .6 giving an area of 27.
Using 5 as the height of the 2 top rectangles would have lengths of 9
and 5, this results in rect 15 having a length of 5 and a height of 3.
This results in Rect X having a length of 9 and a height of 3 giving
an area of 27. The third solution of 6 and 4.5 was given in my
previous answer.Using a height of 20 gives lengths of 2.25 and 1.25.
Dividing 15 by length 1.25 gives a height of 12 which gives Rect X a
length of 2.25 and a height of 12 There may be more solutions if
Rectangles 45 and 25 have additional rational common factors.
Method
2
The value for x is 27.
Label the segments as follows. Top a and c, vertical segments b and
d. a(b) = 45, c(b) = 25, c(d) = 15,and a(d) = x. I will use
substitution to solve. b= 45/a and c = 15/d therefore (45/a)(15/d)=
25. Multiplying both sides by (ad) you get 675 = 25ad . Dividing by
25 gives the answer of 27.
Method
3
x equals 27.
________9__________5_____
| | |
| | |
5| 45 | 25 |5
| | |
|______________|________|
| | |
3| x=27 | 15 |3
| | |
|______________|________|
9 5
The way I figured this problem out was:
I started with 25. I knew that if each side had to be whole numbers,
that the dimensions of the 25 rectangle must be 5 and 5. If you put
the 25 rectangle together with the 15 rectangle, then the short sides
are 5 each, since the opposite sides are congruent. Then in the 15
rectangle, the other side is 3, since 15 divided by 5 is 3. Then in
the 15 + x rectangle, the shorter sides are 3. In the 25 + 45
rectangle, the shorter sides are both 5. In the 45 rectangle alone,
the other side is 9, because 45 divided by 5 is 9. So, in the 45 + x
rectangle, both the shorter sides are 9. Now, in the x rectangle
alone, the dimensions are 3 and 9. 3 times 9 is 27, so x equals 27.
______0.9__________0.5___
| | |
| | |
50| 45 | 25 |50
| | |
|______________|________|
| | |
30| x=27 | 15 |30
| | |
|______________|________|
0.9 0.5
Now, to check my answer, I substituted the whole numbers and used some
decimals. 0.5 times 50 is still 25, o.5 times 30 is still 15, 50
times 0.9 is still 45, so 30 times 0.9 still equals 27.