Download Lesson 1 - Black homework

Unit 5: Lesson 1 Black Homework
Complete the following questions from Signpost in your notebook.
Section
11:01 (# 8, 9, 10)
11:02 (#12)
11:03 (all blue questions)
Answer the following questions in your notebook.
1. How many numbers are less than or equal to 50 are multiples of 3 or 4, but not 5.
2. The sum of three consecutive numbers is a multiple of 89 and is a value of between 600 and
900. What is the sum of the three numbers?
3. What is the least multiple of 72 that has exactly 16 positive factors?
4. Ben has Golden Delicious apples; each weighs 0.6 pounds, and Jonathan apples, each of which
weighs 0.8 pounds. He wants to make apple sauce such that 1/3 of the weight is from
Golden Delicious apples and 2/3 of the weight is from Jonathan apples. He wants to use
all 12 of his Golden Delicious apples. How many Jonathan apples should he use?
5. Using the following 3 clues, can you figure out what number I am?
(1) If I am not a multiple of 4, then I am between 60 and 69
(2) If I am a multiple of 3, I am between 50 and 59
(3) If I am not a multiple of 6, I am between 70 and 79.
What number am I?
6. How many multiples of 7 are between 30 and 790?
7. What is the sum of all the multiples of 3 between 100 and 200?
8. Given:
5
n
m n p m
, what is the value of p?



13 39
156
104
9. What is the largest multiple of 12 that can be written using each digit, 0, 1, 2, 3, 4, 5, 6, 7,
8, and 9 exactly once?
10.
The greatest common factor of two numbers is 14, and the least common multiple is 168.
If the two numbers are not 14 and 168, what is their sum?
Black Homework Solutions
1.
There are 16 multiples of 3 less than 50 and 12 multiples of 4. We have counted the multiples of 12 twice, however, so
we must subtract 4. That is 16 + 12 – 4 = 24. Now we must exclude any multiples of 5 that might have been included.
That would be 3 multiples of 15 (15, 30, and 45) and 2 multiples of 20 (20 and 40). That leaves 24 – 3 – 2 = 19 numbers
less than 50 that are multiples of 3 or 4 but not of 5.
2.
There are four multiples of 89 between 600 and 900: 7 x 89 = 623, 8 x 89 = 712, 9 x 89 =801, and 10 x 89 = 890. The
sum of any three consecutive integers is always a multiple of 3. Since 89 is prime and 9 is the only multiple of 3 in our
list, the sum of the three consecutive integers can only be 801. Therefore the three consecutive integers is are 266,
267, and 268.
3.
We know that 72 = 23 x 32, which means that there are 4 x 3 = 12 factors. If the multiple of 72 included just one
other prime factor different from 2 and 3, it would follow that this multiple m could be represented by m = (23 x 32) x
p, and it would have 4 x 3 x 2 = 24 factors. Bringing in a new prime number creates too many factors for the multiple.
Therefore let’s try multiple (23 x 32) x 2 = 24 x 32, which has 5 x 3 = 15 factors. This doesn’t satisfy our condition, so
we’re left with trying (23 x 32) x 3 = 23 x 33, which has 4 x 4 = 16 factors. Our multiple is 23 x 33 = 216.
4.
The least common multiple of the weights of the two kinds of apples is 2.4 pounds, which means that 4 Golden Delicious
apples weigh the same as 3 Jonathan apples. A ratio of Golden Delicious to 6 Jonathans would give the 1/3 to 2/3
weight ratio that Ben wants for his apple sauce. Since Ben plans to use all 12 of his Golden Delicious, he will need to
use 18 Jonathan apples.
5.
Let’s assume that the mystery number is a multiple of 3, so it must be 51, 54, or 57 according to the second statement.
However it would not 51, 54, or 57 because then it would not be a multiple of 4, according to the first statement it
must be between 60 and 69. So we know that the mystery number is not a multiple of 3. Because it is not a multiple of
3 it also cannot be a multiple of 6 either, so it must be 71, 73, 74, 75, 76, 77 or 79 according to the 3rd statement.
This means that the mystery number is a multiple of 4, because if it were not a multiple of 4 then it would be between
60 and 69. Of the numbers between 70 and 79 that are not multiples of 6, 76 is the only number that is a multiple of 4.
The mystery number is 76.
6.
The first multiple of 7 greater than 30 is 7 x 5 = 35. We could list all the multiples of 7 from 35 to 790 but that would
take a long time! Dividing 790 gives us a quotient of 112 remainder 6, so we know that 7 x 112 < 790 and 7 x 113 > 790.
We know that multiplying any integer from 5 to 112 will give us a multiple of 7 within the desired range. So how many
integers are between 5 through 112? Be careful not to just take the difference between 112 and 5 or you will be one
integer short! We want to include all of the first 112 integers except for the first 4. This means that there are 112 – 4
= 108 integers from 5 to 112, and therefore 108 multiples of 7 between 30 and 790.
7.
Adding the first and last multiples of 3 between 100 and 200, we get 102 + 198 = 300. The second and second-to-last
multiples of 3 also have a sum of 300. Of the 33 multiples of 3 between 100 and 200, we can make 16 pairs that have a
sum of 300. Then add the number 150 that is alone in the middle. The sum is 16 x 300 + 150 = 4950.
8.
Multiplying through by the LCM of 13, 39, 156 and 104 (which is 312), we obtain: 120 = 8n = 2(m + n) = 3(p – m). Solving
120 = 8n for n, we get n = 15. Substituting n = 15 into 120 = 2(m + n), we get 120 = 2(m + 15) & 60 = m + 15 & m = 45.
Using m = 45 in the equation 120 = 3(p – m), we get 120 = 3(p – 45) & 40 = p – 45 & p = 85.
9.
9, 876, 543, 120. Any number that uses all the digits exactly once is divisible by 3. We also need it to be divisible by 4.
If the number named by the last two digits is divisible by 4, then so is the number. So we rearrange the digits in order
but reverse the positions of the 1 and 2 so that the last two digits name a number divisible by 4.
10. Since the least common multiple of the two numbers is 168, both numbers must be factors of 168. If we list all factor
pairs, we get: 168 x 1, 84 x 2, 56 x 3, 42 x 4, 28 x 6, 24 x 7, 21 x 8 and 14 x 12. Since the highest common factor of
the two numbers is 14, the unknown numbers must be multiples of 14. From the list of factors above, the only numbers
that are also multiples of 14 are 84, 56, 42 and 28 (remember we do not include 14 or 168 as possible answers). Of
these four numbers, the two that have a GCF of 14 and an LCM of 168 are 56 and 42. Therefore their sum is 108.