Difficulty: Medium
Correct Answer: 4
Explanation:
Introduction:
This question tests your understanding of how to use the highest common factor (HCF) when a divisor leaves the same remainder in several divisions. This is a standard trick used in many aptitude questions involving remainders and common divisors.
Given Data / Assumptions:
Concept / Approach:
If a number N divides several numbers leaving the same remainder in each case, then N will exactly divide the difference between any pair of those numbers. Therefore, we first compute the pairwise differences and then find the HCF of those differences. That HCF is the greatest possible N.
Step-by-Step Solution:
Compute differences: 4665 − 1305 = 3360 6905 − 4665 = 2240 6905 − 1305 = 5600 Now N must divide 3360, 2240, and 5600 exactly. Find HCF of 3360 and 2240: HCF(3360, 2240) = 1120 Now find HCF of 1120 and 5600: HCF(1120, 5600) = 1120 Therefore N = 1120 Sum of digits of N = 1 + 1 + 2 + 0 = 4
Verification / Alternative check:
To verify, we can calculate the remainders when 1305, 4665, and 6905 are divided by 1120. In each case, the remainder is the same. Also, any larger number would fail to divide at least one of the differences 3360, 2240, or 5600 exactly, so 1120 is indeed the greatest such number.
Why Other Options Are Wrong:
Options 6, 8, and 9 are just different possible digit sums but do not match the actual digit sum of 1120. They would correspond to different hypothetical divisors, which do not satisfy the requirement of being the greatest number that leaves the same remainder in all three divisions.
Common Pitfalls:
A common error is to try to find a divisor of the given numbers directly instead of looking at their differences. Another mistake is to compute only one difference and take its factors without checking the other differences. Always use all given numbers to ensure the divisor works for every pair.
Final Answer:
The greatest number N has digit sum equal to 4.
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