course Mth 151
april 10 around 7:00
assignment 23QA
q001:
the divisors for 2=1,2
3=1,3
4=1,2,4
5=1,5
6=1,2,3,6
7=1,7
8=1,2,4,8
9=1,3,9
10=1,2,5,10
11=1,11
12=1,2,3,4,6,12
13=1,13
14=1,2,7,14
15=1,3,5,15
16=1,2,4,8,16
17=1,17
18=1,2,3,6,9,18
19=1,19
20=1,2,4,5,10,20
*3
q002:
the prime numbers are 2,3,5,7,11,13,17,19
*3
q003:
the prime numbers would be 23,29,31,37
*3
q004:
the twins would be 5,7; 11,13; 17,19
*3
q005:
7 would be the largest number because all the other options would be eliminated due to the process in the problem.
*2
q006:
2,3,5,7 because 4 and 6 are already eliminated
*2
q007:
no because it is divisible with 7
*3
q008:
63:9*7
3*3*7
36:9*4
3*3*4
3*3*2*2
58:2*29
*2
q009:
The numbers would always be prime because we keep having to break down the numbers which are not prime.
*3
q010:
We start by dividing by 3. We get 819 = 3 * 273 = 3 * 3 * 91.
Since 91 isn't divisible by 3, we need to 5 and 7, after which the next prime is 11 and which we will not need to try since 11 * 11 > 91.
91 isn't divisible by 5 since it doesn't end in 0 or 5, but it is divisible by 7 with quotient 13. So 91 = 7 * 13.
Thus we have 819 = 3 * 3 * 91 = 3 * 3 * 7 * 13.
In addition to 1 and 819, the factors of 819 will include
3, 7, 13, all of the prime factors,
3 * 3 = 9, 3 * 7 = 21, 3 * 13 = 39 and 7 * 13 = 91, all of the possible products of two of the prime factors;
and 3 * 3 * 7 = 63, 3 * 3 * 13 = 117, and 3 * 7 * 13 = 273, all of the possible products of three of the prime factors.
q011:
the factors would be: 1,2,3,4,6,8,14,21,12,28,42,24,56,84
*2
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