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The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, is
  • a)
     1677
  • b)
     1683
  • c)
     2523
  • d)
     3363
Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The least number which when divided by 5, 6, 7 and 8 leaves a remainde...
 L.C.M of 5, 6, 7, 8 = 840
Therefore, Required Number is of the form 840k+3.
Least value of k for which (840k+3) is divisible by 9 is k = 2 
Therefore, Required  Number = (840 x 2+3)=1683
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Most Upvoted Answer
The least number which when divided by 5, 6, 7 and 8 leaves a remainde...
Number= LCM of5,6,7&8+3=1683
5++6+8+3=18
which is divisible by 9 hence no reminder
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Community Answer
The least number which when divided by 5, 6, 7 and 8 leaves a remainde...
Understanding the Problem:
The problem states that the least number when divided by 5, 6, 7, and 8 leaves a remainder of 3, but when divided by 9 leaves no remainder.

Solution:
To find the least number that satisfies these conditions, we need to consider the LCM (Least Common Multiple) of 5, 6, 7, and 8 first.

Finding LCM:
- LCM of 5, 6, 7, and 8 = 840

Finding Number:
- Let x be the number we are looking for
- According to the problem, x leaves a remainder of 3 when divided by 5, 6, 7, and 8
- This can be represented as x = 840k + 3, where k is an integer
- Now, x should leave no remainder when divided by 9
- This implies that x should be a multiple of 9
- Substitute x = 840k + 3 into x being a multiple of 9, we get 840k + 3 ≡ 0 (mod 9)
- Solving this congruence, we find k = 2
- Substitute k = 2 back into x = 840k + 3, we get x = 1683
Therefore, the least number that satisfies the given conditions is 1683, which is option B.
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The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder, isa)1677b)1683c)2523d)3363Correct answer is option 'B'. Can you explain this answer?
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