Download Homework 12 Solutions

(2) (3) The answer is correct, so the grade should be 10 points
(5) The answer is correct, so the grade should be 10 points
(7) The rectangle drawn has 12 blocks and 11 is prime not composite. I
would give the student 2 points since it looks like they miscounted the
number of blocks in their rectangle.
(3) (a)
(b)
(c)
(d)
(e)
(f)
The factors of 9 are 3 and 3.
The factors of 16 are 2, 2, 2, and 2.
51 = 3 × 17, so the factors of 51 are 3 and 17.
63 = 7 × 9, so the factors of 63 are 3, 3, and 7.
125 = 5 × 25, so the factors of 125 are 5, 5, and 5.
144 = 12 × 12, and the factors of 12 are 2, 2, and 3, so the factors of 144
are 3, 3, 3, 3, 2, 2, 2, and 2.
(5) (a)
(b)
(c)
(d)
(e)
(f)
7 is prime
3 is prime
9 = 3 × 3 so 9 is composite
12 = 3 × 4, so 12 is composite.
14 = 2 × 7, so 14 is composite
11 is prime.
(13) The top of the tree is 210. 210 = 2 × 105, so 105 goes in the next blank.
105 = 3 × 35 so 35 goes in the final blank. It is possible to find the top number
using only the given numbers because the given numbers are the factors of
the top number and 5 × 7 × 2 × 3 = 210
(19) (a) 419 is prime. To find this, I checked to see if 419 was divisible by any
primes up to 23. Since 232 = 529, I know that there can’t be any factors
of 419 greater than 23. So 419 is prime.
(b) 363 is composite. The sum of the digits is 12 so it is divisible by 3. In
fact 363 = 3 × 121 and 121 = 11 × 11. So 363 = 3 × 11 × 11.
(c) 1369 is composite. 1369 = 37 × 37. This is also its prime factorization,
since 37 is prime.
(d) 1319 is prime. To find this, I checked to see if 1319 was divisible by any
primes up to 37. Since 372 = 1369, I know that there can’t be any factors
of 1319 greater than 37. So 1319 is prime.
(e) 19019 is composite. 19019 = 7 × 2717. 2717 = 11 × 247. 247 = 13 × 19.
So the prime factorization of 19019 is 7 × 11 × 13 × 19.
(f) 15827 is composite. 15827 = 7 × 2261. 2261 = 7 × 323. 323 = 17 × 19.
So the prime factorization of 15827 is 7 × 7 × 17 × 19.
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(23) (a)
(b)
(c)
(d)
This is false. 2 is prime, but 2 is even.
This is false. 5 is a multiple of 5 since 5 = 5 × 1, but 5 is prime.
The statement is true. 2, 3, and 5 are prime and 2 × 3 × 5 = 30
This is false. You would need to check for prime factors less than 13 since
132 = 169 > 127
(29) (a) There are 5 green rods in a one color train representing 15 since there
are 5 threes in 15.
(b) You could make a one color train representing 30 by using 30 of the one
length rods, or you could use 15 of the two length rods, or you could use
10 of the three length rods, or you could use 6 of the five length rods,
or you could use 5 of the six length rods, or you could use 3 of the ten
length rods.
(c) If you can make a one color train that represents the whole number n,
the the number that corresponds to the color of your train is a factor of
n.
(42) The student didn’t complete their factor tree since 9 is composite. The need
to factor 9 as 3 × 3. Then they’ll have the correct factorization of 23 · 33 .