Download Solutions to Test 2 Practice Questions 1. Show how to perform 74

Solutions to Test 2 Practice Questions
1. Show how to perform 74 − 58 and 62 − 35 using the equal additions method.
Adding 2 to both numbers gives
74 − 58 = 76 − 60 = 16.
For the other one, adding five to both sides gives
62 − 35 = 67 − 40 = 27.
The easiest way to justify this is to appeal to the number line model
of subtraction: The difference 74−58 is the length of the gap between
74 and 58 on the number line. Adding two to both numbers simply shifts
this gap to the right two units, without changing its length.
2. Using additive or multiplicative compensation to compute 94 + 88, 37 + 48, 25 × 23 × 4,
and 20 × 35.
There are multiple ways to do compensation to these problems.
include:
94 + 88 = 92 + 90 = 182,
Possibilities
37 + 48 = 35 + 50 = 85,
25 × 23 × 4 = 100 × 23 × 1 = 2300,
20 × 35 = 10 × 70 = 700.
Justification for these uses the associative properties of addition
and multiplication. For example,
20 × 35 = (10 × 2) × 35 = 10 × (2 × 35) = 10 × 70.
3. Compute 18 × 7 and 99 × 15 using the distributive property.
There are multiple ways to do these problems.
Possibilities include:
18 × 7 = (20 − 2) × 7 = (20 × 7) − (2 × 7) = 140 − 14 = 106,
99 × 15 = (100 − 1) × 15 = (100 × 15) − (1 × 15) = 1500 − 15 = 1485.
4. Compute 568 + 493 using expanded notation, and then using any other intermediate
algorithm.
Expanded notation gives
(500 + 60 + 8) + (400 + 90 + 3) = 900 + 150 + 11 = 1061.
One intermediate algorithm is
568
+493
---900
150
+11
---1061
5. Compute 174 + 259 using base-10 blocks.
6. Compute 27 × 13 using base-10 blocks.
7. Compute 65 × 28 using partial products.
65
×28
--1200
480
100
+40
--1820
8. Show how to factor 48 using a factor tree.
9. Explain how the fundamental theorem of arithmetic tells you that when a number is
divisible by 2 and by 3, it is also divisible by 6.
Suppose our number is called N. Since 2 divides N, one can factor
N in such a way that 2 is one of the factors. Similarly, since 3 divides
N, one can factor N in such a way that 3 is one of the factors. But
the FTA tells us that these two factorizations have the same primes
in them. Thus they both have both 2 and 3. It follows that 6 divides
N.
10. Find the greatest common factor of 28 and 35 using the set intersection method and
using the prime factorization method.
Set intersection:
28 :
1, 2, 4, 7, 14, 28.
35 :
1, 5, 7, 35.
So gcf(28, 35) = 7.
Prime factorization:
28 = 2 × 2 × 7,
35 = 5 × 7,
so gcf(28, 35) = 7.
11. Find the least common multiple of 8 and 20 using the set intersection method and using
the prime factorization method.
Set intersection:
8:
8, 16, 24, 32, 40, 48, . . .
20 :
20, 40, 60, . . .
So lcm(8, 20) = 40.
Prime factorization:
8 = 2 × 2 × 2,
so lcm(8, 20) = 2 × 2 × 2 × 5 = 40.
20 = 2 × 2 × 5,