MAKE A TABLEOR CHART

We make a table or chart when we have information that needs to be organized before we can move onand find the solution to a problem. We also use this strategy when we need to find a pattern and generalizeit before we can answer a question. One thing to note here is that the more difficult strategies will sometimes blendtogether and also use the less difficult strategies in their solution. Here is an example of a problem where usinga table or chart would be helpful: Find the greatest common factor of 24, 56, and 108. To do this problem efficiently, we will break each number down into its prime factors, in order from smallest to largest.

24 2, 2, 2, 3

56 2, 2, 2, 7

108 2, 2, 3, 3, 3

After we list these out, we find all the numbers that appear onall three lists. In this case, they all have a first 2 and a second 2.They do not all share any other numbers. So, the greatestcommon factor is 2 x 2, which is 4.

USE A TABLE OR CHART TO SOLVE THE FOLLOWING PROBLEMS

1) WHAT IS THE ONES’ DIGIT OF ?

2) HOW MANY DIAGONALS DOES A REGULAR DECAGON HAVE?

3) FIND THE SMALLEST POSITIVE INTEGER THAT IS DIVISIBLE BY 2, 3, 5, 7, 12, 15, 21, AND 24.

4) IF 1! = 1, 2! = 1X2, 3! = 1X2X3, 4! = 1X2X3X4, ETC. FIND THE ONES’ DIGIT OF 89!

5) IT COSTS 75 CENTS TO USE A COUNTRY TOLL ROAD. THE TOLL MACHINE TAKES EXACT CHANGE ONLY AND ACCEPTS NICKELS, DIMES, AND QUARTERS. HOW MANY COMBINATIONS OF COINS MUST THE MACHINE BE PROGRAMMED TO ACCEPT?

WHAT IS THE ONES’ DIGIT OF ?

= 7 Ones = 7

= 49 Ones = 9

= 343 Ones = 3

= 2401 Ones = 1

= 16807 Ones = 7

= 117649 Ones = 9

= 823543 Ones = 3

= 5764801 Ones = 1

= 403533607 Ones = 7

= 282475249 Ones = 9

AFTER WE DO THIS ENOUGH TIMES TO SEE A PATTERN EMERGE, WE FIND THAT THE ONE’S DIGIT CYCLES EVERY 4 TIMES, GIVING 7, 9, 3, OR 1, RESPECTIVELY. SO, IF WE WANT THE ONES’ DIGIT OF THE 2015TH POWER, WE JUST NEED TO DIVIDE 2015 BY 4, NOTE THE REMAINDER, AND APPLY IT TO FIND THE SOLUTION. 2015 / 4 = 503 R 3. SINCE THE REMAINDER IS 3, THIS FITS WITH THE THIRD NUMBER IN THE CYCLE, WHICH IS 3. NOTE: IF WE HAVE NO REMAINDER, THAT IS THE SAME AS GETTING A REMAINDER OF 4 AND OUR CHOICE WOULD BE 1.

HOW MANY DIAGONALS DOES A REGULAR DECAGON HAVE?

TO SOLVE THIS PROBLEM, WE WOULD ALSO NEED TO DRAW THE FIGURES, CONNECT THE DIAGONALS, AND COUNT THEM. THAT IS NOT SHOWN HERE DUE TO SPACE CONSTRAINTS. HOWEVER, FROM THE CHART TO THE RIGHT, WE CAN SEE A PATTERN EMERGING. THE PATTERN IS 0, 2, 5, 9. THIS PATTERN IS: +2. +3, +4, +5, ETC. TO GET TO 10 SIDES, WE NEED TO CONTINUE THIS PATTERN OUT FOR 4 MORE TIMES. HERE IS OUR COMPLETED PATTERN: 0, 2, 5, 9, 14, 20, 27, 35. SO, A DECAGON WOULD HAVE 35 DIAGONALS.

1 side Does not make a figure

2 sides Does not make a figure

3 sides 0

4 sides 2

5 sides 5

6 sides 9

FIND THE SMALLEST POSITIVE INTEGER THAT IS DIVISIBLE BY 2, 3, 5, 7, 12, 15, 21,

AND 24.

BEFORE WE START MAKING OUR TABLE OR CHART, WE CAN LOOK MORE CLOSELY AT THE PROBLEM TO MAKE IT SIMPLER FOR US. (I CALL THIS WORKING SMART AS OPPOSED TO WORKING HARD.) SINCE 2 GOES INTO 12, WE CAN REMOVE 2 AS A CHOICE. THIS IS ALSO TRUE FOR 3 (INTO 12), 5 (INTO 15), 7 (INTO 21) AND 12 (INTO 24). HMMM . . . IS IT OK TO REMOVE 12 SINCE WE SAID 2 AND 3 WENT INTO THEM? THE ANSWER IS YES, BECAUSE 2 AND 3 ALSO GO INTO 24. HERE IS OUR TABLE AND SOLUTION.

15 Prime factors = 3,5

21 Prime factors = 3,7

24 Prime factors = 2,2,2,3

We need a least common multiple here, so all of the above numbers must be represented. Going from the smallest to largest prime factor, we would need 3 2’s, one 3, one 5, and one 7. This would give us 2x2x2x3x5x7. Take a minute and check to see that all of the above numbers go into this, and then compute the answer, which is 840.

IF 1! = 1, 2! = 1X2, 3! = 1X2X3, 4! = 1X2X3X4, ETC. FIND THE ONES’ DIGIT OF

89!

I ESPECIALLY ENJOY THIS PROBLEM BECAUSE THERE IS AN INTERESTING TWIST IN IT. IF WE LOOK AT THE CHART ON THE RIGHT, WE SEE THAT WHEN WE SOLVE FOR 5! AND BEYOND, THE ONES’ DIGIT IS ALWAYS 0. THEREFORE, THE ANSWER IS 0.

1! = 1 Ones’ digit = 1

2! = 2 Ones’ digit = 2

3! = 6 Ones’ digit = 6

4! = 24 Ones’ digit = 4

5! = 120 Ones’ digit = 0

6! = 720 Ones’ digit = 0

7! = 5040 Ones’ digit = 0

8! = 40320 Ones’ digit = 0

IT COSTS 75 CENTS TO USE A COUNTRY TOLL ROAD. THE TOLL MACHINE TAKES EXACT CHANGE ONLY AND ACCEPTS NICKELS, DIMES, AND QUARTERS. HOW MANY

COMBINATIONS OF COINS MUST THE MACHINE BE PROGRAMMED TO ACCEPT?

Q D N Q D N

3 0 0 1 0 10

2 2 1 0 7 1

2 1 3 0 6 3

2 0 5 0 5 5

1 5 0 0 4 7

1 4 2 0 3 9

1 3 4 0 2 11

1 2 6 0 1 13

1 1 8 0 0 15

THIS PROBLEM USES THE ‘LIST THE POSSIBILITIES’ STRATEGY EXCEPT THAT IT IS MORE COMPLEX AND WILL REQUIRE A SYSTEMATIC WAY TO ARRIVE AT THE SOLUTION. I SOLVED THIS BY (A) USING THE MOST QUARTERS I COULD AND THEN (B) USING THE MOST DIMES I COULD. THE NICKELS ADJUSTED THEMSELVES ACCORDINGLY. FROM THE TABLE, WE CAN SEE THAT THE MACHINE MUST BE PROGRAMMED TO ACCEPT 18 COMBINATIONS.

MORE TABLE AND CHART PROBLEMS AND RESOURCES ONLINE

• HTTP://WWW.CK12.ORG/BOOK/CK-12-ALGEBRA-I-SECOND-EDITION/R17/SECTION/1.8/

• HTTPS://WWW.TEACHERVISION.COM/MATH/PROBLEM-SOLVING/48897.HTML

• HTTPS://WWW.YOUTUBE.COM/WATCH?V=M_2P06RSMPK

• HTTPS://WWW.YOUTUBE.COM/WATCH?V=V0UGEDMGSQS

• HTTPS://WWW.BRAINPOP.COM/MATH/DATAANALYSIS/PROBLEMSOLVINGUSINGTABLES/PREVIEW.WEML

• HTTP://WWW.PORTANGELESSCHOOLS.ORG/STUDENTS/GRADE-4-ST.HTML