1. TRY THE POSSIBILITIES

2. Try the possibilities to discover the answer that works. 1)Find a number between 400 and 450 that is divisible by 2, 3, 5, and 7. 2)When the children visited the farm and saw all the goats and chickens, they counted 39 heads and 108 legs. How many were goats and how many were chickens? 3)Find the smallest prime number that is larger than 300. 4)A perfect number is one in which the number itself is equal to the sum of its factors (excluding the number itself). One example of a perfect number is 28. The factors of 28 are 1, 2, 4, 7, 14, and 28. If we remove 28 and add the other factors, we get 1 + 2 + 4 + 7 + 14, which equals 28. Find the only one digit perfect number. 5)Maria invited three other couples to a dinner party. After she puts the invitations in the mailbox, she realized she didn’t check to see if the right invitations were placed in the right envelopes. How many ways could she have put them in the envelopes so that at least one person got the right invitation?

3. 1) FIND A NUMBER BETWEEN 400 AND 450 THAT IS DIVISIBLE BY 2, 3, 5, AND 7. ONE SOLUTION OF MANY: WE NEED A NUMBER BETWEEN 400 AND 450, BUT THIS GIVES US 50 NUMBERS TO TRY. WE CAN CUT THIS DOWN BY LOOKING AT THE NUMBERS WE ARE TESTING. IN ORDER FOR 2 TO DIVIDE THE NUMBER, IT MUST BE EVEN; AND IN ORDER FOR 5 TO DIVIDE THE NUMBER, IT MUST END IN 0 OR 5. SINCE THE NUMBER MUST BE EVEN, THOUGH, IT MUST END IN 0. OUR POSSIBILITIES NOW BECOME 400, 410, 420, 430, 440, AND 450. NOW WE JUST CHECK TO SEE WHICH OF THESE CAN BE DIVIDED BY 3 AND 7. 3 GOES INTO 420 AND 450. OF THESE TWO CHOICES, ONLY 420 IS DIVISIBLE BY 7. THEREFORE, THE ANSWER IS 420.

4. 2) WHEN THE CHILDREN VISITED THE FARM AND SAW ALL THE GOATS AND CHICKENS, THEY COUNTED 39 HEADS AND 108 LEGS. HOW MANY WERE GOATS AND HOW MANY WERE CHICKENS? WE COULD SOLVE THIS USING A SYSTEM OF EQUATIONS, BUT THIS PROCESS MAY BE BEYOND OUR STUDENTS. IN THAT CASE, WE CAN TRY THE POSSIBILITIES UNTIL WE FIND THE ONE THAT FITS. THERE WILL BE MANY WAYS TO GO ABOUT THIS, BUT HERE IS ONE OF THEM. WE KNOW WE HAVE 39 HEADS, SO THE NUMBER OF GOATS + THE NUMBER OF CHICKENS = 39. I WILL BEGIN BY LOOKING AT 20 GOATS AND 19 CHICKENS. THIS WILL GIVE 118 LEGS. THIS IS TOO MANY, SO WE WILL TAKE THE NUMBER OF GOATS DOWN TO 19. 19 GOATS AND 20 CHICKENS WILL GIVE 116 LEGS. NEXT I WILL TRY 17 GOATS AND 22 CHICKENS. THIS GIVES 112 LEGS. NOW I WILL TRY 15 GOATS AND 24 CHICKENS. THIS GIVES 108 LEGS, WHICH IS WHAT WE ARE LOOKING FOR. THE ANSWER IS 15 GOATS AND 24 CHICKENS.

5. 3) FIND THE SMALLEST PRIME NUMBER THAT IS LARGER THAN 300. IN ORDER FOR A NUMBER TO BE PRIME, THE ONLY FACTORS IT HAS ARE 1 AND ITSELF. SO, TO CUT DOWN THE POSSIBILITIES, WE CAN REMOVE ANY EVEN NUMBER (2 WILL GO INTO IT) AND ANY NUMBER THAT ENDS IN 5 (5 WILL GO INTO IT). OUR FIRST POSSIBILITY IS 301. 3 WILL NOT DIVIDE THIS NUMBER, BUT 7 WILL, SO IT IS NOT PRIME. NEXT, WE WILL TRY 303. 3 DIVIDES THIS NUMBER, SO THAT IS NOT PRIME. THE NEXT NUMBER WE WILL TRY IS 307. 3, 7, 11, 13, AND 17 WILL NOT DIVIDE THIS NUMBER, SO IT IS PRIME. OUR ANSWER IS 307.

6. 4) A PERFECT NUMBER IS ONE IN WHICH THE NUMBER ITSELF IS EQUAL TO THE SUM OF ITS FACTORS (EXCLUDING THE NUMBER ITSELF). ONE EXAMPLE OF A PERFECT NUMBER IS 28. THE FACTORS OF 28 ARE 1, 2, 4, 7, 14, AND 28. IF WE REMOVE 28 AND ADD THE OTHER FACTORS, WE GET 1 + 2 + 4 + 7 + 14, WHICH EQUALS 28. FIND THE ONLY ONE DIGIT PERFECT NUMBER. HERE IS ONE SOLUTION: I WILL BEGIN WITH THE NUMBER 1 UNTIL I FIND THE NUMBER THAT IS PERFECT. 1.THE ONLY FACTOR FOR 1 IS 1. IF WE REMOVE THAT, THE SUM IS 0. NOT THIS ONE 2.THE FACTORS FOR 2 ARE 1 AND 2. IF WE REMOVE 2, THE SUM IS 1. NOT THIS ONE EITHER 3.THE FACTORS FOR 3 ARE 1 AND 3. IF WE REMOVE 3, THE SUM IS 1. NOPE 4.THE FACTORS FOR 4 ARE 1, 2, AND 4. IF WE REMOVE 4, THE SUM IS 3. WRONG AGAIN 5.THE FACTORS FOR 5 ARE 1 AND 5. IF WE REMOVE 5, THE SUM IS 1. NOT YET 6.THE FACTORS FOR 6 ARE 1, 2, 3, AND 6. IF WE REMOVE 6, THE SUM IS 6. BINGO! SO, THE ANSWER FOR THIS PROBLEM IS 6.

7. 5) Maria invited three other couples to a dinner party. After she puts the invitations in the mailbox, she realized she didn’t check to see if the right invitations were placed in the right envelopes. How many ways could she have put them in the envelopes so that at least one person got the right invitation? HERE IS ONE WAY TO SOLVE THE PROBLEM. CALL THE OTHER COUPLES A, B, AND C. NOW, LIST ALL THE WAYS MARIA COULD HAVE STUFFED THE ENVELOPES. HERE WE GO! THE WAY THEY SHOULD HAVE BEEN STUFFED: A, B, C POSSIBILITIES A, B, C (3 RIGHT) A, C, B (1 RIGHT) B, A, C (1 RIGHT) B, C, A (0 RIGHT) C, A, B (0 RIGHT) C, B, A (1 RIGHT) OF THE 6 TOTAL POSSIBILITIES, 4 OF THEM GIVE AT LEAST 1 RIGHT INVITATION. SO, THE ANSWER IS 4 WAYS.

8. MORE TRY THE POSSIBILITY PROBLEMS AND RESOURCES ONLINE HTTP://WWW.STUDYZONE.ORG/TESTPREP/MATH4/D/GUESSCHECK4L.CFM HTTPS://WWW.YOUTUBE.COM/WATCH?V=VWTWXE2DQUA HTTPS://WWW.TEACHERVISION.COM/MATH/PROBLEM-SOLVING/48896.HTML HTTPS://WWW.YOUTUBE.COM/WATCH?V=WIPYQUPV9WY HTTP://WWW.IXL.COM/MATH/GRADE-5/GUESS-AND-CHECK-PROBLEMS HTTPS://WWW.YOUTUBE.COM/WATCH?V=BSRCT7TCIO8 HTTP://PRED.BOUN.EDU.TR/PS/PS3.HTML

9. Note: this is the first Page of a 3 page document on how to solve problems using the trial and error method. It is a part of the AMP Project through the University of Akron under the direction of Dr. Antonio Quesada. To locate this document, follow these steps: 1.Go to www.google.comwww.google.com 2.Type in the phrase ‘math problem solving trial and error’ 3.Look for and click on the tab that says: [DOC]Solving Problems by Trial and Error ur... [DOC]Solving Problems by Trial and ErrorSolving Problems by Trial and Error