Download Math 1350 Review #1

Math 1350 Review #3(answers)
1. Write the following using a single exponent:
45
7
12
a) 3  3
b) 8
2
5
2
45  2 
210
7
12
19


 22
3 3  3
8
8
8
2
2
2
2. Solve for x in the following:
c) 712  212 143
712  212  143   7  2   143
12
 1412  143  1415
b)  3x   320
4
a) 37  3x  313
7  x  13
4 x  20
x 6
x 5
c) 3x  2 x  65 x8
x  5x  8
4 x  8
x 2
3. Find a whole number greater than 1 and less than 100 that is both a square and a cube.
number  be , where e is a multiple of 2, and e is a multiple of 3. So e is a multiple of 6.
b
b6
b12
2 64 4096
3 729
4. Solve the following problems by completing the intermediate algorithm:
a) 376  594
b) 56  73
56
376
 73
 594
18  3  6 
10 sum of ones
160 sum of tens
800 sum of hundreds
970 final sum
150  3  50 
420  70  6 
3500  70  50 
4088 final product
5. Solve the following problems by completing the Lattice Method:
a) 568  493
b) 37  196
5
+ 4
1 0
9
0
6
9
1
1
5
6
1061
3
8
3
0
1
7 2
1
3
7
7
0
7
6
3
2 1 8 42
5
2
7252
1
9
6
6. Find the product 27  51 by completing the Egyptian Method:
27
51
1
51
2
102
4
204
8
408
16
816
1  2  8  16  27 , so 27  51  51  102  408  816  1377
7. Find the product 37  48 by completing the Russian Peasant Method:
Halving Doubling
37
48
18
96
9
192
4
384
2
768
1
1536
37  48  48  192  1536  1776
8. Find the quotient and remainder of 7261  43 by completing the following intermediate
algorithm:
8
160
43 7261 How many 43’s in 7261?
6880 Guess: 160
381
344
How many 43’s in 381?
Guess: 8
37
7261  43  168 r 37
9. a)Complete the following base six addition table:
+ 0 1 2 3
4
4
0 0 1 2 3
5
1 1 2 3 4
2 2 3 4 5 10
3 3 4 5 10 11
4 4 5 10 11 12
5 5 10 11 12 13
5
5
10
11
12
13
14
b) Complete the following Lattice Method calculation of 45six  23six :
4
5 six
3six
+ 2
1 1
0
1
1
2
2 six
112 six
c) Complete the following base six multiplication table:
5
 0 1 2 3 4
0
0
0 0 0 0 0
4
5
1 0 1 2 3
2 0 2 4 10 12 14
3 0 3 10 13 20 23
4 0 4 12 20 24 32
5 0 5 14 23 32 41
d) Complete the following intermediate algorithm calculation of 123six  43six
123six
 43six
13 3  3
100 3  20
300 3  100
200 40  3
1200 40  20
4000 40  100
10213 final product
10. Test the following numbers for divisibility by 2, 3, 4, 5, 6, 7, 8, 9, 10, and 11:
Number
2 3 4 5 6 7 8 9 10 11
1234567891020 Y Y Y Y Y N N N Y
N
1234567891096 Y N Y N N N Y N N
N
1234567891029 N Y N N N N N N N
Y
3
6
12
Y Y Y Y Y N Y Y Y
N
2 3 5
11. Factor the following numbers into primes:
a)
10,800
100
20
10
5
108
5
2
2
54
2
2
27
3
9
3
3
10,800  24  33  52
b)
3289
11
299
23
13
3289  11  13  23
12. What is the smallest counting number divisible by 2, 4, 6, 8, 10, 12, and 14?
22
4  22
6  23
8  23
10  2  5
12  2 2  3
14  2  7
Every multiple of all these numbers must have 23  3  5  7  840 as a factor.
13. The primes 2 and 5 differ by 3. Show why this is the only pair of primes whose difference
is 3.
Suppose that there are two prime numbers p1 and p2 with p2  p1  3 and p1  2 . Since all
prime numbers except 2 are odd numbers, this means that p2 would have to be a sum of
odd numbers. But a sum of odd numbers must be even. This makes p2 an even prime
number larger than 2, which is impossible. So 2 and 5 are the only prime numbers that
differ by 3.