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Find the number lying between 900 and 1000 which when divided by 38 and 57 leaves in each case a remainder 23. (SSC Multitasking 2014)
- a)912
- b)926
- c)935
- d)962
Correct answer is option 'C'. Can you explain this answer?
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Find the number lying between 900 and 1000 which when divided by 38 an...
L.C.M of (38, 57) = 114
Multiple of 114 between 900 and 1000 = 912
number which leaves 23 = 912 + 23 = 935
Multiple of 114 between 900 and 1000 = 912
number which leaves 23 = 912 + 23 = 935
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Find the number lying between 900 and 1000 which when divided by 38 an...
Problem Overview
To find a number between 900 and 1000 that, when divided by 38 and 57, leaves a remainder of 23.
Formulating the Equation
Let the required number be \( x \). According to the problem, we can express this as:
- \( x \equiv 23 \mod{38} \)
- \( x \equiv 23 \mod{57} \)
This implies:
- \( x - 23 \) is divisible by both 38 and 57.
Finding the Least Common Multiple (LCM)
To solve for \( x \), we need the LCM of 38 and 57.
- The prime factorization of 38 is \( 2 \times 19 \).
- The prime factorization of 57 is \( 3 \times 19 \).
The LCM is calculated by taking the highest power of each prime:
- \( \text{LCM}(38, 57) = 2^1 \times 3^1 \times 19^1 = 114 \).
Setting Up the Equation
Since \( x - 23 \) is divisible by 114, we can write:
- \( x - 23 = 114k \) for some integer \( k \).
Thus,
- \( x = 114k + 23 \).
Finding Suitable Values of \( k \)
We need to find \( k \) such that:
- \( 900 < 114k="" +="" 23="" />< 1000="" />
Rearranging the inequalities:
- \( 877 < 114k="" />< 977="" />
Dividing by 114 gives:
- \( 7.68 < k="" />< 8.57="" />
Since \( k \) must be an integer, the only possible value is \( k = 8 \).
Calculating \( x \)
Substituting \( k = 8 \):
- \( x = 114 \times 8 + 23 = 912 + 23 = 935 \).
Conclusion
The number that lies between 900 and 1000 and meets the criteria is \( \textbf{935} \). Thus, the correct answer is option 'C'.
To find a number between 900 and 1000 that, when divided by 38 and 57, leaves a remainder of 23.
Formulating the Equation
Let the required number be \( x \). According to the problem, we can express this as:
- \( x \equiv 23 \mod{38} \)
- \( x \equiv 23 \mod{57} \)
This implies:
- \( x - 23 \) is divisible by both 38 and 57.
Finding the Least Common Multiple (LCM)
To solve for \( x \), we need the LCM of 38 and 57.
- The prime factorization of 38 is \( 2 \times 19 \).
- The prime factorization of 57 is \( 3 \times 19 \).
The LCM is calculated by taking the highest power of each prime:
- \( \text{LCM}(38, 57) = 2^1 \times 3^1 \times 19^1 = 114 \).
Setting Up the Equation
Since \( x - 23 \) is divisible by 114, we can write:
- \( x - 23 = 114k \) for some integer \( k \).
Thus,
- \( x = 114k + 23 \).
Finding Suitable Values of \( k \)
We need to find \( k \) such that:
- \( 900 < 114k="" +="" 23="" />< 1000="" />
Rearranging the inequalities:
- \( 877 < 114k="" />< 977="" />
Dividing by 114 gives:
- \( 7.68 < k="" />< 8.57="" />
Since \( k \) must be an integer, the only possible value is \( k = 8 \).
Calculating \( x \)
Substituting \( k = 8 \):
- \( x = 114 \times 8 + 23 = 912 + 23 = 935 \).
Conclusion
The number that lies between 900 and 1000 and meets the criteria is \( \textbf{935} \). Thus, the correct answer is option 'C'.
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Find the number lying between 900 and 1000 which when divided by 38 and 57 leaves in each case a remainder 23. (SSC Multitasking 2014)a)912b)926c)935d)962Correct answer is option 'C'. Can you explain this answer?
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