Divisor Calculation for Numbers with the Same Remainder

The problem asks us to find the greatest number, let's call it x, that divides 955, 1027, and 1075, leaving the same remainder in each case.

When a number x divides several numbers and leaves the same remainder, it means that the differences between these numbers must be exactly divisible by x.

Let the numbers be $N_1 = 955$, $N_2 = 1027$, and $N_3 = 1075$. Let the common remainder be $r$. We can write: $N_1 = q_1 \times x + r$ $N_2 = q_2 \times x + r$ $N_3 = q_3 \times x + r$ where $q_1, q_2, q_3$ are the quotients.

Now, let's find the differences between these numbers:

• Difference 1: $N_2 - N_1 = 1027 - 955 = 72$
• Difference 2: $N_3 - N_2 = 1075 - 1027 = 48$
• Difference 3: $N_3 - N_1 = 1075 - 955 = 120$

The greatest number x must be the Greatest Common Divisor (GCD) of these differences (72, 48, and 120). We can find the GCD using prime factorization.

First, find the prime factors of each difference:

• Prime factorization of $72$: $72 = 8 \times 9 = (2^3) \times (3^2) = 2 \times 2 \times 2 \times 3 \times 3$
• Prime factorization of $48$: $48 = 16 \times 3 = (2^4) \times 3 = 2 \times 2 \times 2 \times 2 \times 3$
• Prime factorization of $120$: $120 = 12 \times 10 = (2^2 \times 3) \times (2 \times 5) = 2^3 \times 3 \times 5 = 2 \times 2 \times 2 \times 3 \times 5$

To find the GCD, we take the lowest power of each common prime factor.

The common prime factors are 2 and 3. The lowest power of 2 common to all is $2^3$. The lowest power of 3 common to all is $3^1$.

So, the GCD(72, 48, 120) = $2^3 \times 3^1 = 8 \times 3 = 24$.

Therefore, the greatest number x is 24.

We can quickly verify this by checking the remainder when 24 divides the original numbers: $955 \div 24 = 39$ remainder $19$ ($955 = 24 \times 39 + 19$) $1027 \div 24 = 42$ remainder $19$ ($1027 = 24 \times 42 + 19$) $1075 \div 24 = 44$ remainder $19$ ($1075 = 24 \times 44 + 19$) The remainder is 19 in all cases, confirming that $x = 24$.

Non-Factor Identification for the Divisor x

The question asks which of the given options is NOT a factor of $x$, where $x = 24$. The options are 6, 8, 4, and 16. Let's check each option to see if it divides 24 without leaving a remainder.

• Option 1: 6 Is 6 a factor of 24? Calculation: $24 \div 6 = 4$. Since the result is an integer, 6 is a factor of 24.
• Option 2: 8 Is 8 a factor of 24? Calculation: $24 \div 8 = 3$. Since the result is an integer, 8 is a factor of 24.
• Option 3: 4 Is 4 a factor of 24? Calculation: $24 \div 4 = 6$. Since the result is an integer, 4 is a factor of 24.
• Option 4: 16 Is 16 a factor of 24? Calculation: $24 \div 16 = 1.5$. Since the result is not an integer, 16 is NOT a factor of 24.

Based on these checks, the number 16 is the only option that does not divide 24 evenly.