Computers, Data and Programming

Running head: ASSESSMENT ITEM 2
Assignment 1: Computers, Data and Programming
Name of the Student
Name of the University
Author’s Note

1
ASSESSMENT ITEM 2
Question 1 Answer
a. The value of base b if (152)b = 0x6A.
Demonstration of the calculation with the following steps:
0x6A = (6A)₁₆ = 6 * 16¹ + A*16 =⁰ 96 + 10 = 106 (A = 10 in Hexadecimal)
(152)b = 1 * b² + 5 * b¹ + 2 * b⁰
=> (152)b = b² + 5b + 2
Comparing both the equation we get:
=> b² + 5b + 2 = 106
=> b² + 5b - 104 = 0
=> b² + 13b - 8b - 104 = 0
=> b(b+13) - 8(b + 13) = 0
=> (b-8) (b+13) = 0
b = 8 ( b ≠ - 13 as base cannot be negative)
so b = 8
b. Conversion of the following
i. 0xBAD into 3-base representation
Step#1: Translating into decimal:
0xBAD16 = 11∙162+10∙161+13∙160 = 2816+160+13 = 298910
Result: 298910
Converting 298910 in 3-ary system here so:

2
ASSESSMENT ITEM 2
The result is obtained by dividing the number with its base
2989 3
-
2988
996 3
1
-
996
332 3
0
-
330
110 3
2
-
108
36 3
2 -36 12 3
0 -12 4 3
0 -3 1
1
Conversion Result:
BAD16 = 110022013
ii. 3217 into 2-base (binary) representation
3217 = 3∙72+2∙71+1∙70 = 147+14+1 = 16210
Result: 16210
Converting 16210 in Binary system here so: