Understanding the Problem: Counting Ones in Binary

The question asks us to find the total count of the digit '1' in the binary representation of a given mathematical expression: $3 \times 512 + 5 \times 64 + 7 \times 8 + 3$. To solve this, we first need to evaluate the expression and then convert the result into its binary form to count the '1's.

Evaluating the Mathematical Expression

The expression involves powers of 2, which is helpful when thinking about binary representation. Let's rewrite the expression using powers of 2:

• $512 = 2^9$
• $64 = 2^6$
• $8 = 2^3$
• $3 = 3 \times 2^0$

So the expression is: $3 \times 2^9 + 5 \times 2^6 + 7 \times 2^3 + 3 \times 2^0$.

Let's calculate the value of each term:

• $3 \times 512 = 1536$
• $5 \times 64 = 320$
• $7 \times 8 = 56$
• $3 \times 1 = 3$

Now, we sum these values:

$1536 + 320 + 56 + 3 = 1915$.

So, we need to find the number of '1's in the binary representation of 1915.

Converting 1915 to Binary Representation

We can convert 1915 to binary using repeated division by 2, keeping track of the remainders.

Reading the remainders from bottom to top, the binary representation of 1915 is $11101111011_2$.

Counting the Number of 1s in Binary

Now, let's count the occurrences of the digit '1' in the binary string $11101111011_2$:

• $1$ (most significant bit)
• $1$
• $1$
• $0$
• $1$
• $1$
• $1$
• $1$
• $0$
• $1$
• $1$ (least significant bit)

Counting the '1's, we get $1 + 1 + 1 + 0 + 1 + 1 + 1 + 1 + 0 + 1 + 1 = 9$.

Alternatively, we can analyze the expression $3 \times 2^9 + 5 \times 2^6 + 7 \times 2^3 + 3 \times 2^0$ directly in binary. Since the powers of 2 ($2^9, 2^6, 2^3, 2^0$) are spaced sufficiently far apart, the binary representations of the individual terms $c \times 2^p$ will not have overlapping '1' bits.

Let's look at the binary representation of the coefficients:

• $3 = 11_2$ (has two 1s)
• $5 = 101_2$ (has two 1s)
• $7 = 111_2$ (has three 1s)
• $3 = 11_2$ (has two 1s)

When a number $c_2$ is multiplied by $2^p$, its binary representation is shifted left by $p$ positions. The number of '1's remains the same. Since the powers (9, 6, 3, 0) differ by at least 3, and the largest coefficient (7) requires only 3 bits ($111_2$), the '1's from one term $c_i \times 2^{p_i}$ will not overlap with the '1's from another term $c_j \times 2^{p_j}$.

• $3 \times 2^9$: $11_2$ shifted left by 9 positions. '1's at bit positions $0+9=9$ and $1+9=10$. (Two 1s)
• $5 \times 2^6$: $101_2$ shifted left by 6 positions. '1's at bit positions $0+6=6$ and $2+6=8$. (Two 1s)
• $7 \times 2^3$: $111_2$ shifted left by 3 positions. '1's at bit positions $0+3=3$, $1+3=4$, and $2+3=5$. (Three 1s)
• $3 \times 2^0$: $11_2$ shifted left by 0 positions. '1's at bit positions $0+0=0$ and $1+0=1$. (Two 1s)

The positions with '1's are {10, 9}, {8, 6}, {5, 4, 3}, {1, 0}. Since these sets of positions are disjoint, the total number of '1's is simply the sum of the number of '1's in each coefficient:

Total number of 1s = (Number of 1s in 3) + (Number of 1s in 5) + (Number of 1s in 7) + (Number of 1s in 3)

Total number of 1s = $2 + 2 + 3 + 2 = 9$.

Both methods yield the same result.

Final Count of Digit 1

The number of digit 1 present in the binary representation is 9.

Revision Table: Key Concepts

Additional Information: Binary Arithmetic and Bit Manipulation

Understanding binary representation is fundamental in computer science and digital electronics. The expression given in the question highlights how numbers can be represented as a sum of terms involving powers of 2. When the coefficients are also small integers, their binary representations determine the pattern of '1's at specific positions determined by the associated power of 2.

Multiplying a binary number by $2^n$ is equivalent to shifting its binary representation $n$ places to the left and filling the new low-order bits with zeros. For example, $11_2 \times 2^3 = 11000_2$. The number of '1's remains the same during a left shift.

Adding binary numbers involves standard column addition with carries. However, in this specific problem, because the powers of 2 ($2^9, 2^6, 2^3, 2^0$) are relatively far apart compared to the number of bits in the coefficients ($3=11_2$, $5=101_2$, $7=111_2$), the '1' bits from each term fall into distinct position ranges, preventing overlaps (carries across term boundaries). This allows for the shortcut of summing the number of '1's in the binary representation of each coefficient.

For instance:

• $3 \times 2^9$ adds '1's at bit positions 9 and 10.
• $5 \times 2^6$ adds '1's at bit positions 6 and 8.
• $7 \times 2^3$ adds '1's at bit positions 3, 4, and 5.
• $3 \times 2^0$ adds '1's at bit positions 0 and 1.

The set of bit positions {0, 1, 3, 4, 5, 6, 8, 9, 10} contains all the positions where a '1' appears in the final binary number. The total number of '1's is the count of unique positions in this combined set, which is 9.