Find the greatest number that will divide 964, 1238, and 1400 leaving remainders of 41, 31, and 51 respectively in each case. Use the idea of equal remainders and highest common factor to determine this largest divisor.

Introduction / Context:
This is another example of a remainder based highest common factor (HCF) question. When a number divides several integers and leaves potentially different remainders, we can transform the problem by subtracting those remainders and turning it into a pure divisibility question. Then the largest such divisor is the HCF of the adjusted values.

Given Data / Assumptions:

• A number N divides 964 leaving remainder 41.
• The same N divides 1238 leaving remainder 31.
• The same N divides 1400 leaving remainder 51.
• We want the greatest such N.

Concept / Approach:
If N divides several numbers and leaves remainders r1, r2, r3, the quantities (number - remainder) for each case are exact multiples of N. Thus, N must be a common divisor of these adjusted values. The greatest such N is the highest common factor of those adjusted numbers. So we compute the three adjusted numbers and then find their HCF.

Step-by-Step Solution:
Compute adjusted values: 964 - 41 = 923 1238 - 31 = 1207 1400 - 51 = 1349 Now find HCF(923, 1207, 1349). First compute HCF of 923 and 1207. Using Euclidean method, HCF(923, 1207) = 71. Now HCF(71, 1349) is also 71. Therefore, N = 71.

Verification / Alternative check:
Check each original number with N = 71: 964 / 71 = 13 remainder 41. 1238 / 71 = 17 remainder 31. 1400 / 71 = 19 remainder 51. All remainders match the given values, confirming that 71 is a valid divisor. Since it comes from the HCF of the adjusted values, it is the greatest such divisor.

Why Other Options Are Wrong:
64, 58, 69, and 82 either do not divide all the adjusted values exactly or produce different remainders than those specified in the problem, so they cannot be the required greatest number.

Common Pitfalls:
A typical error is to try to work directly with the original numbers and remainders without forming the adjusted numbers by subtraction. Another mistake is to compute HCF using only two of the adjusted numbers and to forget to include the third, which can lead to a divisor that is not common to all three cases.

Final Answer:
71