Difficulty: Medium
Correct Answer: 4
Explanation:
Introduction / Context:
This question uses a standard trick in number theory. When a number divides several integers leaving the same remainder each time, the differences between those integers are all exactly divisible by that number. This converts a remainder based question into a highest common factor (HCF) problem, which can then be solved systematically.
Given Data / Assumptions:
Concept / Approach:
If a number N divides several integers leaving the same remainder r, then:
(4665 - 1305), (6905 - 4665), and (6905 - 1305)
are all multiples of N. Therefore, N must be a common divisor of these differences, and the greatest such N is the HCF of these differences. Once N is found, we just add its digits for the final answer.
Step-by-Step Solution:
Compute differences:
4665 - 1305 = 3360
6905 - 4665 = 2240
6905 - 1305 = 5600
Find HCF of 3360, 2240, and 5600.
First, HCF(3360, 2240) = 1120.
Then HCF(1120, 5600) = 1120.
So N = 1120.
Sum of digits of N = 1 + 1 + 2 + 0 = 4.
Verification / Alternative check:
We can check that 1120 divides all differences:
3360 / 1120 = 3, 2240 / 1120 = 2, 5600 / 1120 = 5.
All results are integers, confirming that 1120 is a common divisor. Since we used the HCF, it is the greatest such number, and the digit sum 4 is therefore correct.
Why Other Options Are Wrong:
6, 7, 8, and 9 do not represent the sum of digits of the correct greatest number 1120. They might correspond to digit sums of smaller divisors, but the question specifically asks for the greatest one.
Common Pitfalls:
A common error is to try to find N directly from the original numbers without using differences, or to confuse the idea with LCM. Another mistake is to take the difference of only one pair and not compute the HCF of all three differences, which might result in a smaller divisor rather than the greatest one.
Final Answer:
4
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