August 2021 – Physicspupil

Problem 1 of Project Euler

If we list all the natural numbers below $10$ that are multiples of $3$ or $5$ , we get $3$ , $5$ , $6$ and $9$ . The sum of these multiples is $23$ .

/* running a check on the first two statements by a direct computation */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);
    system("pause");
    return 0;
    }

/* running a check on the first two statements by a direct computation */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

printf("The sum of all the natural numbers below 10 that are multiples of 3 and 5 is %d.", 3 + 5 + 6 + 9);

system("pause");

return 0;

}

Find the sum of all the multiples of $3$ or $5$ below $1000$ .

Ans. $233168$

(step-by-step approach, slow but sure logic)

I wish to find all the multiples of $3$ (excluding multiples of $5$ ) below $1000$ :

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 3)
        {
        if (i % 5 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);
    }

/* finding all the multiples of 3 (excluding multiples of 5) below 1000*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 3)

{

if (i % 5 != 0)

{

sum += i;

}

printf("The sum of multiples of three (excluding multiples of five) below one thousand is %d.", sum);

}

Output.
The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

1 2	Output. The sum of multiples of three (excluding multiples of five) below one thousand is 133668.

Then I wish to find all the multiples of $5$ (excluding multiples of $3$ ) below $1000$ :

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 5)
        {
        if (i % 3 != 0)
            {
            sum += i;
            }
        }
    printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);
    }

/* finding all the multiples of 5 (excluding multiples of 3) below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 5)

{

if (i % 3 != 0)

{

sum += i;

}

printf("The sum of multiples of five (excluding multiples of three) below one thousand is %d.", sum);

}

Output.
The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

1 2	Output. The sum of multiples of five (excluding multiples of three) below one thousand is 66335.

Lastly I wish to find all the multiples of $15=3\times 5$ below $1000$ :

/* finding all the multiples of 15 below 1000:*/
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    for (int i = 0; i < 1000; i += 15)
        {
        sum += i;
        }
    printf("The sum of multiples of fifteen below one thousand is %d.", sum);
    }

/* finding all the multiples of 15 below 1000:*/

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

for (int i = 0; i < 1000; i += 15)

{

sum += i;

}

printf("The sum of multiples of fifteen below one thousand is %d.", sum);

}

Output.
The sum of multiples of fifteen below one thousand is 33165.

1 2	Output. The sum of multiples of fifteen below one thousand is 33165.

In sum,

$\begin{aligned} & \quad \textrm{The sum of all multiples of 3 or 5 below 1000} \\ & = 133\small,668 + 66\small,335 + 33\small,165 \\ & = 233\small,168 \end{aligned}$

Ans. $233168 \quad\checkmark$

(to be continued)

Going off at a tangent, the sum of all numbers from $1$ to $1000$ is $500500$ . See:

#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int i,sum;
    i = 1;
    sum = 0;
    do
    {
    sum += i++;
    }
    while (i <= 1000);
    printf("sum = %d", sum);
    system("pause");
    return 0;
    }

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int i,sum;

i = 1;

sum = 0;

{

sum += i++;

}

while (i <= 1000);

printf("sum = %d", sum);

system("pause");

return 0;

}

Output.
sum = 500500

1 2	Output. sum = 500500

(proof)

$\begin{aligned} & \quad 1+2+3+\cdots +1000 \\ & = \sum_{k=1}^{1000}k \\ & = \frac{(1000)(1000+1)}{2} \\ & = 500500 \\ \end{aligned}$

(continue)

Solution. (fast and furious attempt)

/* The sum of all the multiples of 3 or 5 below 1000 */
#include <stdio.h>;
#include <stdlib.h>;
int main()
    {
    int sum = 0;
    int i = 0;
    while(i < 1000)
    {
    if(i % 3 == 0 || i % 5 == 0)
    sum += i;
    i++;
    }
    printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);
    }

/* The sum of all the multiples of 3 or 5 below 1000 */

#include <stdio.h>;

#include <stdlib.h>;

int main()

{

int sum = 0;

int i = 0;

while(i < 1000)

{

if(i % 3 == 0 || i % 5 == 0)

sum += i;

i++;

}

printf("The sum of all the multiples of 3 or 5 below 1000 is %d.", sum);

}

Output.
The sum of all the multiples of 3 or 5 below 1000 is 233168.

1 2	Output. The sum of all the multiples of 3 or 5 below 1000 is 233168.

Given two operands $a$ and $b$ , let them be integers:

int a;
int b;
printf("Please input two integers \n==>");
scanf("%d %d", &a, &b);

int a;

int b;

printf("Please input two integers \n==>");

scanf("%d %d", &a, &b);

Addition Operator $(+)$ adds two operands $a$ , $b$ :

sum = a + b;
printf("The sum is %d\n", sum);

1 2	sum = a + b; printf("The sum is %d\n", sum);

Subtraction Operator $(-)$ subtracts the second operand $b$ from the first operand $a$ :

difference = a - b;
printf("The difference is %d\n", difference);

1 2	difference = a - b; printf("The difference is %d\n", difference);

Multiplication Operator $(*)$ multiplies both operands $a$ , $b$ :

product = a * b;
printf("The product is %d\n", product);

1 2	product = a * b; printf("The product is %d\n", product);

Division Operator $(/)$ divides the first operand $a$ (numerator) by the second operand $b$ (denominator):

quotient = a / b;
printf("The quotient is %d\n", quotient);

1 2	quotient = a / b; printf("The quotient is %d\n", quotient);

Modulo Operator $(\% )$ returns the remainder after integer division of the first operand $a$ (dividend) by the second operand $b$ (divisor):

s = a % b;
printf("a mod b is %d\n", s);

1 2	s = a % b; printf("a mod b is %d\n", s);

Increment Operator $(++)$ increases the integer value by one.

int a;
int b;
//pre-increment
b = ++a;
//post-increment
b = a++;

int a;

int b;

//pre-increment

b = ++a;

//post-increment

b = a++;

Decrement Operator $(--)$ decreases the integer value by one.

int a;
int b;
//pre-decrement
b = --a;
//post-decrement
b = a--;

int a;

int b;

//pre-decrement

b = --a;

//post-decrement

b = a--;

Cf.

The pre-increment and pre-decrement operators increment (decrement resp.) their operand by 1, and the value of the expression is the resulting incremented (decremented resp.) value. The post-increment and post-decrement operators increase (decrease resp.) the value of their operand by 1, but the value of the expression is the operand’s value prior to the increment (decrement resp.) operation.

_{Wikipedia on Increment and decrement operators}

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