Answer:
// here is code in c++.
#include <bits/stdc++.h>
using namespace std;
// main function
int main() {
// variable
int num;
int tot_sum=0;
cout<<"enter a number: ";
// read the value of n
cin>>num;
// check the multiple of 2 or 5 from 1 to n
for(int x=1;x<=num;x++)
{
// if multiple of 2 or 5 then add them
if(x%2==0 ||x%5==0)
{
tot_sum=tot_sum+x;
}
}
// print the sum
cout<<"sum of multiple of 2 or 5 is:"<<tot_sum<<endl;
return 0;
}
Explanation:
Read the number "n" from user.Check every number from 1 to n, if it is a multiple of 2 or 5 then add them to "tot_sum".When the for loop end, "tot_sum" will have sum of all the number which are either multiple of 2 or 5.Print the sum.
Output:
enter a number:11
sum of multiple of 2 or 5 is:35
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
