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If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs?
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If counted in pairs,one will remain..If counted in threes, two will re...
To find the number of eggs in the basket, we need to solve a series of modular equations based on the conditions provided. Let's break down the problem:
Understanding the Conditions
- If counted in pairs (2), one will remain:
**Equation:** \( x \equiv 1 \mod 2 \)
- If counted in threes (3), two will remain:
**Equation:** \( x \equiv 2 \mod 3 \)
- If counted in fours (4), three will remain:
**Equation:** \( x \equiv 3 \mod 4 \)
- If counted in fives (5), four will remain:
**Equation:** \( x \equiv 4 \mod 5 \)
- If counted in sixes (6), five will remain:
**Equation:** \( x \equiv 5 \mod 6 \)
- If counted in sevens (7), nothing will remain:
**Equation:** \( x \equiv 0 \mod 7 \)
Finding a Common Solution
These conditions can be rephrased as follows:
- \( x = 2k + 1 \) for some integer \( k \) (from the first condition).
- \( x = 3m + 2 \) (second condition).
- \( x = 4n + 3 \) (third condition).
- \( x = 5p + 4 \) (fourth condition).
- \( x = 6q + 5 \) (fifth condition).
- \( x = 7r \) (sixth condition).
By analyzing the equations, we see that for the first five conditions, \( x \) is always one less than the base number (2, 3, 4, 5, 6). Thus, \( x + 1 \) must be a multiple of 2, 3, 4, 5, and 6. The least common multiple (LCM) of these numbers is 60. Therefore:
- \( x + 1 = 60k \) for some integer \( k \)
- Thus, \( x = 60k - 1 \)
Considering the Last Condition
For the seventh condition \( x \equiv 0 \mod 7 \):
- Substitute \( x = 60k - 1 \) into this equation:
\( 60k - 1 \equiv 0 \mod 7 \)
or \( 60k \equiv 1 \mod 7 \).
- Calculate \( 60 \mod 7 \):
\( 60 \equiv 4 \mod 7 \). Thus:
\( 4k \equiv 1 \mod 7 \).
The multiplicative inverse of 4 modulo 7 is 2, so multiply both sides by 2:
- \( k \equiv 2 \mod 7 \)
- Therefore, \( k = 7m + 2 \) for some integer \( m \).
Final Calculation
Substitute back to find \( x \):
- \( x = 60(7m + 2) - 1 = 420m + 119 \).
To satisfy
Understanding the Conditions
- If counted in pairs (2), one will remain:
**Equation:** \( x \equiv 1 \mod 2 \)
- If counted in threes (3), two will remain:
**Equation:** \( x \equiv 2 \mod 3 \)
- If counted in fours (4), three will remain:
**Equation:** \( x \equiv 3 \mod 4 \)
- If counted in fives (5), four will remain:
**Equation:** \( x \equiv 4 \mod 5 \)
- If counted in sixes (6), five will remain:
**Equation:** \( x \equiv 5 \mod 6 \)
- If counted in sevens (7), nothing will remain:
**Equation:** \( x \equiv 0 \mod 7 \)
Finding a Common Solution
These conditions can be rephrased as follows:
- \( x = 2k + 1 \) for some integer \( k \) (from the first condition).
- \( x = 3m + 2 \) (second condition).
- \( x = 4n + 3 \) (third condition).
- \( x = 5p + 4 \) (fourth condition).
- \( x = 6q + 5 \) (fifth condition).
- \( x = 7r \) (sixth condition).
By analyzing the equations, we see that for the first five conditions, \( x \) is always one less than the base number (2, 3, 4, 5, 6). Thus, \( x + 1 \) must be a multiple of 2, 3, 4, 5, and 6. The least common multiple (LCM) of these numbers is 60. Therefore:
- \( x + 1 = 60k \) for some integer \( k \)
- Thus, \( x = 60k - 1 \)
Considering the Last Condition
For the seventh condition \( x \equiv 0 \mod 7 \):
- Substitute \( x = 60k - 1 \) into this equation:
\( 60k - 1 \equiv 0 \mod 7 \)
or \( 60k \equiv 1 \mod 7 \).
- Calculate \( 60 \mod 7 \):
\( 60 \equiv 4 \mod 7 \). Thus:
\( 4k \equiv 1 \mod 7 \).
The multiplicative inverse of 4 modulo 7 is 2, so multiply both sides by 2:
- \( k \equiv 2 \mod 7 \)
- Therefore, \( k = 7m + 2 \) for some integer \( m \).
Final Calculation
Substitute back to find \( x \):
- \( x = 60(7m + 2) - 1 = 420m + 119 \).
To satisfy
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If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs?
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If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs?.
If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs? for UPSC 2025 is part of UPSC preparation. The Question and answers have been prepared according to the UPSC exam syllabus. Information about If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs? covers all topics & solutions for UPSC 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for If counted in pairs,one will remain..If counted in threes, two will remain..If counted in fours,three will remain..If counted in fives, four will remain..If counted in sixes,five will remain..If counted in sevens, nothing will remain..My basket cannot accomodate more than 150 eggs?.
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