# Efficient way to multiply with 7

We can multiply a number by 7 using bitwise operator. First left shift the number by 3 bits (you will get 8n) then subtract the original numberfrom the shifted number and return the difference (8n – n).

Program:

## C

```# include<stdio.h>

int multiplyBySeven(unsigned int n)
{
/* Note the inner bracket here. This is needed
because precedence of '-' operator is higher
than '<<' */
return ((n<<3) - n);
}

/* Driver program to test above function */
int main()
{
unsigned int n = 4;
printf("%u", multiplyBySeven(n));

getchar();
return 0;
}
```

## Python

```# Python program to multiply any
# positive number to 7

# Function to mutiply any number with 7
def multiplyBySeven(n):

# Note the inner bracket here.
# This is needed because
# precedence of '-' operator is
# higher than '<<'
return ((n << 3) - n)

# Driver code
n = 4
print(multiplyBySeven(n))

# This code is contributed by Danish Raza
```

Output:
`28`

Time Complexity: O(1)
Space Complexity: O(1)

Note: Works only for positive integers.
Same concept can be used for fast multiplication by 9 or other numbers.

# GATE CS Corner    Company Wise Coding Practice

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