Finding the Greatest Divisor with Remainders

The problem asks us to find the greatest number that divides both 87 and 123, but not perfectly. Instead, it leaves a remainder of 7 when dividing 87 and a remainder of 3 when dividing 123.

Let the greatest number we are looking for be '$d$'. According to the problem statement:

• When 87 is divided by '$d$', the remainder is 7. This means $87 = q_1 d + 7$, where $q_1$ is the quotient. This implies that '$d$' must divide $(87 - 7)$ perfectly.
• When 123 is divided by '$d$', the remainder is 3. This means $123 = q_2 d + 3$, where $q_2$ is the quotient. This implies that '$d$' must divide $(123 - 3)$ perfectly.

So, we need to find a number '$d$' that divides both $(87 - 7)$ and $(123 - 3)$ perfectly. Let's calculate these values:

• $87 - 7 = 80$
• $123 - 3 = 120$

Now, the problem is reduced to finding the greatest common divisor (GCD) of 80 and 120. The GCD is the largest number that divides both 80 and 120 without leaving any remainder.

Calculating the GCD of 80 and 120

We can find the GCD using a few methods. One common method is using prime factorization.

• Prime factorization of 80: $80 = 8 \times 10 = (2 \times 2 \times 2) \times (2 \times 5) = 2^4 \times 5^1$
• Prime factorization of 120: $120 = 12 \times 10 = (2 \times 2 \times 3) \times (2 \times 5) = 2^3 \times 3^1 \times 5^1$

To find the GCD, we take the lowest power of each common prime factor. The common prime factors are 2 and 5.

• The lowest power of 2 is $2^3$.
• The lowest power of 5 is $5^1$.

So, the GCD(80, 120) is: $GCD(80, 120) = 2^3 \times 5^1 = 8 \times 5 = 40$.

Alternative Method: Euclidean Algorithm

We can also use the Euclidean algorithm to find the GCD of 120 and 80.

1. Divide 120 by 80: $120 = 1 \times 80 + 40$
2. Now divide 80 by the remainder 40: $80 = 2 \times 40 + 0$

The last non-zero remainder is 40. Therefore, the GCD(80, 120) is 40.

Conclusion

The greatest common divisor of 80 and 120 is 40. This means 40 is the greatest number that divides 87 leaving a remainder of 7 ($87 = 2 \times 40 + 7$) and divides 123 leaving a remainder of 3 ($123 = 3 \times 40 + 3$).