Find the greatest number that will divide 964, 1238, and 1400, leaving remainders 41, 31, and 51 respectively. What is this greatest number?

Introduction:
This question is another classic remainder and HCF problem. A single number divides several given numbers but leaves different remainders. We must use the differences between the numbers and their remainders to determine the greatest possible divisor.

Given Data / Assumptions:

• On division by the required number N:
• 964 leaves remainder 41
• 1238 leaves remainder 31
• 1400 leaves remainder 51
• We must find the greatest such N.

Concept / Approach:
If N divides 964 and leaves remainder 41, then 964 − 41 is exactly divisible by N. Similarly for the other numbers. Therefore, N divides the adjusted values 964 − 41, 1238 − 31, and 1400 − 51. These adjusted values are all multiples of N, so N must be a common divisor of them. The greatest such N is the HCF of those adjusted values.

Step-by-Step Solution:
Compute adjusted values: 964 − 41 = 923 1238 − 31 = 1207 1400 − 51 = 1349 Now we need HCF(923, 1207, 1349). Observe divisibility: 923 = 13 × 71 1207 = 17 × 71 1349 = 19 × 71 All three numbers have the common factor 71 and no higher common factor. Therefore HCF(923, 1207, 1349) = 71 Hence, the greatest number that satisfies the given conditions is 71

Verification / Alternative check:
Check by division: 964 ÷ 71 gives quotient 13 and remainder 41. 1238 ÷ 71 gives quotient 17 and remainder 31. 1400 ÷ 71 gives quotient 19 and remainder 51. All remainders match the problem statement, confirming that 71 works. No larger common divisor exists because 923, 1207, and 1349 share only 71 as a common prime factor.

Why Other Options Are Wrong:
64, 69, and 58 do not divide the adjusted values 923, 1207, and 1349 exactly. At least one of the divisions would leave a non zero remainder, which violates the requirement that N must divide each adjusted value perfectly.

Common Pitfalls:
One common mistake is to try to directly find a number that fits the three original remainders without subtracting them. Another error is to compute the HCF of the original numbers 964, 1238, and 1400, which does not incorporate the remainder information and leads to the wrong answer.

Final Answer:
The greatest number that satisfies all the given conditions is 71.