Problem Solving Strategies (continued)

STAR 7

 

5B5 pg157: Comic of the Month

I ) Understanding the Problem

The problem stated that there is a shipping and handing charge.  That is why the final prices for each order are not evenly divisible by the number of magazines he ordered.

 

II) List Clues: Table Form

 

Month Single Order Total Comics Cost ($)
1 5 5 3.07
2 2 7 1.72
3 6 13 3.52
4 3 16 2.17

 

II ) Look for a Pattern

 

–       3.07/ 5 = 0.614

–       1.72/2 = 0.860

–       3.52/6 = 0.587

–       2.17/3 = 0.723

–       Notice how all the results are different.  If there was no shipping and handling, all the quotients would be the same

–       Notice also the price for 2 comics and 3 comics.  They are 1.72 and 2.17 respectively.  If the handling fee is a flat rate, which we are assuming it is, the difference of this number should be the cost of 1 comic.

–       2.17 – 1.72 = 0.45 = Cost of 1 Comic

–       1.72 = 2 Comics and Handling = 0.9 + handling

–       Handling = 0.82

 

III ) Conclusion

 

–       K = Cost (in dollars) = (Number of Comics)(Cost per Comic) + Handling Fee

–       C = number of comics

–       K = C(0.45) + 0.82

–       48(0.45) + 0.82 = 22.42

–       Cost of 48 Comics = $22.42

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Problem Solving Strategies

STAR 1
These STAR problems are designed to help math students understand the thinking process behind solving a math problem.  I chose the following problem to explain first; it was the first problem I had trouble solving at the beginning of my problem solving strategies class at Hunter College. In the end, I used the same method I did at first which is a mix of various problem solving skills but mostly Eliminating Possibilities. The first time around I couldn’t find a solution, but the second time, I kept better track of the numbers I tried already.

HOUSES
+HOTELS
CONTROL

I ) Eliminate possibilities for the far left digit

Based on the way these cryptarithms are solved in Ch.3, we can begin to solve this one in a similar fashion. HOUSES, HOTELS, both six letters, both represent six digit numbers. The largest 6 digit number is 999,999. 999,999+999,999 = 1,999,998.
In these problems, if the left most digit is 0, there would not be a letter there. We can safely assume that C is not 0. We also concluded that the biggest C can be is 1.

1st Conclusion: C = 1

Also take note that HOUSES and HOTELS can’t even be 999,999 since they have different digits throughout the number.

II ) Look for an easy number to start working with. H looks interesting.

H+H > 10
If H = 9, H+H can be at most 19 with a carry from the previous column. Take note that the largest possible carry is 1.

Therefore, H+H+1 > 10, 2H > 9, H > 9/2

2nd Conclusion: H ≥ 5

III ) Are there other numbers we can limit possibilities with?

S+S = L or 1L (as in 1L could equal 10, 11, 12, etc…)
There is no carry from the right so this is certain. Therefore, we know that L is an even digit.

3rd Conclusion: L = even digit

Take note that O and N do not have to be even because there could be a carry from the columns to their right, respectively.
We can also conclude that U has to be 0 or 9 in order for U+T+ (possible carry) = T
Take note: If U = 9 then S+E+ (possible carry) ≥ 10.

4th Conclusion: U = 0 or 9

IV ) Trial and Error: The tedious part.

There’s really no other option at this point. We will have to just try out some numbers.

Where to start?
Try with designating H as a certain number, since H ≥ 5. That limits the number we need to try by a half. Also by choosing an H will limit your possibilities for O, which appears often. It would be useful to find out what O is early on in the problem.

Trail 1: H = 5

We will zoom in on HO + HO + (possible carry) = CON to start each trail.

We know that O can’t be 1 since C = 1, also O+O+(possible carry) would not be greater than or equal to 10. O has to be 0 if H = 5.
50
+50
10N

Now we are left with a contradiction with N
0 + 0 + (possible carry) = 0 or 1 = N
N can’t be 0 since O = 0, N can’t be 1 since C = 1.
Nothing works.

5th Conclusion: H ≠ 5

Trail 2: H = 6

6O
+6O
1ON

This one is not as obvious to see at first. At first glace you might say that O could be 3 from a carry, but notice that O + O + (possible carry) = 7 at most if O = 3. Therefore, there won’t be any carry onto the H column. O = 2, if H = 6.

We can now limit CON = 124 or 125 if there is a carry from U+T.
HOU+HOT = CONT

We need to test them both. Let’s test with the carry first. That means U = 9. If U = 0,
U + T + (possible carry) would only produce a carry if T = 9 and it received a carry. At that point, U + T + (possible carry) ≠ T.

Testing with the assumption that there would be a carry from S+E.
629
+62T
125T

Now let’s look at SES + ELS = ROL, and keep in mind the numbers we already used.
629SES
+62TELS
125TR2L

L = 4 or 8 since 2 and 6 are taken. We can start our test from there.

Try L = 4, S ≠ 2 (taken), so S = 7.
7E7
+E47
R24

Only possible number left for E are 3 and 8.
1 + 4 + (3 or 8) ≠ 2 or 12
L ≠ 4

Try L = 8, S ≠ 9 (taken), S = 4
4E4
+E84
R28

If E+8=12 then E = 4 which is already taken.
L ≠ 8

Therefore, if H = 6, U ≠ 9.

Now try without the carry, as in H = 6, U = 0.
620SES
+62TELS
124TR2L

L = 8 since 2, 4, 6 are taken.
S ≠ 4 (taken), S = 9
9E9
+E89
R28

E = 3 is the only option. Carry + 3 + 8 = 12.
939
+389
428
R = 4 is a contradiction. 4 is already taken.

We have now tried all possibilities involving H = 6.

6th Conclusion: H ≠ 6.

Trail 3: H = 7

Using the same method, we can go through the same process with H = 7. However, in the interest of saving paper, I will just say that after testing all possibilities for H = 7, I was not able to find any numbers that worked. I would like to jump straight to Trail 8 where I found the solution.

7th Conclusion: H ≠ 7

Trail 4: H = 8

86
+86
16(2 or 3)

Is not possible due to the carry from 6+6.
Therefore O = 7 if H = 8.
87
+87
17(4 or 5)

Let’s test with the carry first, as in U has to be 9.
879SES
+87TELS
175TROL

Test L = 2
S ≠ 1 (taken), S = 6
6E6
+E26
R72

For this to work E = 4. 1 + 4 + 2 = 7

9646
+T426
T072
Everything is working so far, and the last number remaining is 3, which also works.
T = 3.

We found numbers that work. Due to the nature of these types of problems, we can assume there is only 1 answer.

V ) Expressing your answer

These are just a couple ways you can express your answer.
Copy form:

HOUSES
+HOTELS
CONTROL

879646
+873426
1753072

List form:
Number 1 2 3 4 5 6 7 8 9 0
Letter C L T E N S O H U R

STAR 3
The rest of the Star Problems is mostly from the text Problem Solving Strategies 2nd Ed, by Ken Johnson and Ted Herr. For the sake of saving paper, I will not write the original problems from the text book. Please refer to the text book while reading this walkthrough. For example, the next problem will be from Chapter 3, Problem Set B, Question 4, page 72. The short hand notation I will use is 3B4 pg72.

3B4 pg72: High Scorers

I ) Understanding the Problem
There are 5 girls on the basketball team. Each of them scored a different score. Using the clues, we should be able to eliminate possibilities for each person’s score until only one possibility remains.

II ) List the Clues
– Seaside Shooters has 5 Starters.
– They scored all the points in the game with a final score of 95-94. For now, let’s assume they scored 95 and won the game since the problem didn’t say anything.
– Kellene: Everyone’s totals were odd
– Sara: Donna is 4th with 17. I [Sara] scored 12 more than Kellene.
– Martina: Kellene and I scored a total of 30 points. I outscored her.
– Heather: The last digit in everyone’s score was different.
– Donna: Our highest scorer had 25 points.

III ) Start Eliminating Possibilities
– Start looking for the most general clue which will help you eliminate the most possibilities
– Look at: They’re all odd, and the highest score is 25. This eliminates all the numbers 26 and up and all the even numbers.
– The 2nd thing I noticed is that Donna is the 4th highest and everyone seems to outscore Kellene. So my next assumption is that Kellene is the lowest scorer.

IV ) Test out Some Numbers
– Reminder of our Givens: Total = 95, Donna = 17, 25 is the highest and all their last digits are unique

Let’s try Sara = 25
– Sara is 12 more than Kellene so Kellene = 25 – 11 = 13
– Martina + Kellene = 30, which means Martina is 17 which is a contradiction. Donna is already 17

Let’s try Sara = 23
– Kellene = 23 – 12 = 11
– Martina = 30 – 11 = 19
– Donna = 17
– This leaves Heather = 25

V ) Conclusion
– Using the process of guessing, eliminating possibilities and trying something new, we came to our conclusion. Heather = 25, Sara = 23, Martina = 19, Donna = 17, and Kellene = 11.

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