Find the largest number that divides 1305, 4665, and 6905 leaving the same remainder in each case. Also determine that common remainder.
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A1210, 158
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B1120, 158
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C1120, 185
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D1210, 185
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ENone of these
Answer
Correct Answer: 1120, 185
Explanation
Introduction / Context:Problems that ask for the largest number dividing several integers with the same remainder are classic Highest Common Factor (HCF) applications. The key insight is that if a number leaves the same remainder upon dividing multiple values, then it must exactly divide all pairwise differences of those values.
Given Data / Assumptions:
- Numbers: 1305, 4665, 6905
- Let the required divisor be D and the common remainder be r.
- Then 1305 = D*k1 + r, 4665 = D*k2 + r, 6905 = D*k3 + r for some integers k1, k2, k3.
Concept / Approach:The same remainder implies D divides pairwise differences. Hence D = HCF of {4665 - 1305, 6905 - 4665, 6905 - 1305}. After D is found, compute r from any original number: r = number - D * floor(number / D).
Step-by-Step Solution:
Compute differences: 4665 - 1305 = 3360 Compute differences: 6905 - 4665 = 2240 Compute differences: 6905 - 1305 = 5600 Find D = HCF(3360, 2240, 5600) HCF(3360, 2240) = 1120; then HCF(1120, 5600) = 1120 ⇒ D = 1120 Common remainder r from 1305: r = 1305 - 1120 = 185Verification / Alternative check:Check with 4665: 4665 / 1120 = 4 remainder 185 (since 1120*4 = 4480; 4665 - 4480 = 185). Check with 6905: 1120*6 = 6720; 6905 - 6720 = 185. Same remainder confirms correctness.
Why Other Options Are Wrong:
- 1210, 158: 1210 does not divide all differences 3360, 2240, 5600.
- 1120, 158: 1120 is correct but r = 185, not 158.
- 1210, 185: 1210 is not the HCF of the differences.
- None of these: A matching correct pair exists, so this is invalid.
Common Pitfalls:Using HCF of the original numbers instead of their differences, or computing the remainder with an incorrect dividend.
Final Answer:Largest divisor = 1120 and common remainder = 185.