Understanding the Problem

We need to find the largest possible integer that, when it divides 152, leaves a remainder of 2, and when it divides 181, leaves a remainder of 6. Let this greatest number be denoted by $ 'd' $.

Adjusting Numbers Based on Remainders

If a number $ 'N' $ is divided by $ 'd' $ and leaves a remainder $ 'r' $, it means that $ N - r $ is perfectly divisible by $ 'd' $. Applying this logic:

• For the number 152 with a remainder of 2, the number perfectly divisible by $ 'd' $ is $ 152 - 2 = 150 $.
• For the number 181 with a remainder of 6, the number perfectly divisible by $ 'd' $ is $ 181 - 6 = 175 $.

Therefore, the number $ 'd' $ we are looking for must be a common divisor of both 150 and 175. Since we want the greatest such number, we need to find the Highest Common Factor (HCF) of 150 and 175.

Calculating the HCF of 150 and 175

We can find the HCF using the prime factorization method:

• Prime factorization of 150:
  • $ 150 = 10 \times 15 $
  • $ 150 = (2 \times 5) \times (3 \times 5) $
  • $ 150 = 2 \times 3 \times 5 \times 5 $
  • $ 150 = 2 \times 3 \times 5^2 $
• Prime factorization of 175:
  • $ 175 = 5 \times 35 $
  • $ 175 = 5 \times (5 \times 7) $
  • $ 175 = 5 \times 5 \times 7 $
  • $ 175 = 5^2 \times 7 $

To find the HCF, we identify the common prime factors raised to the lowest power they appear in either factorization.

The common prime factor is 5. The lowest power it appears with is $ 5^2 $.

So, $ \text{HCF}(150, 175) = 5^2 = 25 $.

Conclusion

The greatest number that divides 152 leaving a remainder of 2 and 181 leaving a remainder of 6 is the HCF of the adjusted numbers (150 and 175), which is 25.