Download Multiplying and Dividing Integers - Black Problems

U1L5 Multiplying and Dividing Integers - Black Problems
1. Screech Names That Number. Screech and Lisa are playing a game called Name That Number. Here is the
target number and the cards they have in this round:
The goal is to use as many of the five cards as you can to form an expression that equals the target
number. You can use addition and subtraction, but not multiplication or division.
Question: Screech was able to make an expression that used all five of the cards. What might his
expression have been?
Extra: Use all five cards and each of the operations +, -, x, / once to form the target number. Grouping
symbols are allowed
2. Guess My Number If You Can! I’m thinking of a number. Put away your calculators. Here are the only
clues I'm going to give you.
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My number when divided by 2 has a remainder of 1.
My number when divided by 3 has a remainder of 2.
My number when divided by 4 has a remainder of 3.
My number when divided by 5 has a remainder of 4.
My number when divided by 6 has a remainder of 5.
My number when divided by 7 has a remainder of 6.
My number when divided by 8 has a remainder of 7.
My number when divided by 9 has a remainder of 8.
Well, maybe one more clue: my number is the smallest integer for which the above clues hold true.
After you find my number, I want you to tell me HOW you can find my number. I will not accept guess
and check or trial and error as the reason for this problem.
I want you to practice using words like:
 division NOT goes into
 multiply NOT timesed
 factor
 product
 quotient
 divisor
 dividend
Good luck. This one may be harder for you to explain than to solve.
3. Alternating Sequence. Consider the infinite sum 96 + 48 + 24 + 12 + …, where each addend is half of the
previous addend and the pattern continues without end. The sum of the first ten terms is 191.8125. Each
addend gets you midway from where you are to 192, so the sum gets as close as you want to 192. Next,
consider the sum 72 + (-36) + 18 + (-9) + 4.5 + …, whose addends are alternatively positive and negative
and where each addend is (-1/2) times the previous addend. This infinite sum is some whole number.
Find that whole number.
4. Maximize a Product. The values of a, b, c, and d are 1, 2, 3, and 4 – but not necessarily in that order. Find
the largest possible value of ab + bc + cd + da.
Multiplying and Dividing Integers - Black Solutions
1. His expression might have been: 5 - (-7) - 3 + (-4) + (-1)
To get my answer, I worked with the different operations and plugged them between the numbers to
see if I could get an answer of four. I discovered that there were many ways to get the answer of four
using just the operations of [+] and [-].
The first one I came up with was
5 - (-7) - 3 + (-4) + (-1)
^ ^ ^ ^
12 9 5 4
The answer is four, which is correct. Therefore, this equation may have been one of Screech’s
equations.
2. The smallest integers for which the clues hold true is 2519. I first tried to find the answer using only the
first 2 clues. I listed several numbers that when divided by 2 have a remainder of 1 and several numbers
that when divided by 3 have a remainder of 2.
2: 3, 5, 7, 9, 11, 13…
3: 5, 8, 11, 14, 17, 20…
The smallest common integer was 5. I then listed the numbers that when divided by 4 have a remainder
of 3 and looked for the smallest common integer for 2, 3, and 4.
4: 7, 11, 15, 19, 23, 27… .The smallest common integer is 11.
I continued this process for one more number to see if I could find a pattern. I had to extend my previous
lists at this point.
For 2:
3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55, 57, 59,
61…
For 3:
5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62…
For 4:
7, 11, 15, 19, 23, 27, 31, 35, 39, 43, 47, 51, 55, 59, 63…
For 5: (integers that when divided by 5 have a remainder of 4)
9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64….
The smallest common integer is 59.
I tried to see a pattern with dividing by 2, 3, 4, and 5 and with their respective smallest integers: 5, 11,
and 59. I came up with the following solution: The least common multiple of 2 and 3 is 6, the least
common multiple of 2, 3 and 4 is 12, and the least common multiple of 2, 3, 4 and 5 is 60. The smallest
integers in each of my 3 answers so far was 1 less than the least common multiple.
I then found the least common multiple of the integers from 2 to 9.
2=2, 3=3, 4=2x2, 5=5, 6=2x3, 7=7, 8=2x2x2, 9=3x3. The least common multiple is 2x2x2x3x3x5x7=2520.
1 less than this number, 2520-1=2519 which is my answer. I divided 2519 by 2, 3, 4, 5, 6, 7, 8, and 9 and
the remainders given in the problem checked.
3. Alternating Sequence. 48. You can try to evaluate the sum when two, three, four, five, six, et cetera
terms are involved and make a guess.
There is formula for the sum of an infinite geometric sequence whose common ratio is, in absolute value,
less than 1. It is S = a/(1 – r), where a is the first term and r is the common ratio. For the given sequence,
a = 72 and r = -1/2. So, S = 72(1 – (-0.5)) = 72/(1.5) = 48.
4. Maximize a Product. 25. Note that ab + bc + cd + da = b(a+c) + d(c+a) = (a+c)(b+d). The possible products
are 3 x 7, 4 x 6, and 5 x 5. The last of these is maximal.