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Find the least number which when divided by 4, 12 and 16 leaves remainder 3 in each case, but it is perfectly divisible by 7.
  • a)
    48
  • b)
    51
  • c)
    99
  • d)
    147
Correct answer is option 'D'. Can you explain this answer?
Verified Answer
Find the least number which when divided by 4, 12 and 16 leaves remain...
L. C. M. of 4, 12, 16 = 48
∴ 48 × 1 + 3 = 51, Not divisible by 7
and 48 × 2 + 3 = 99,
Not divisible by 7 but, 48 × 3 + 3 = 147, Divisible 7
∴ Lowest multiple of 7 = 147
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Most Upvoted Answer
Find the least number which when divided by 4, 12 and 16 leaves remain...
Explanation:

Step 1: Find the LCM of 4, 12, and 16
- LCM of 4, 12, and 16 is 48.

Step 2: Find the number that leaves a remainder of 3 when divided by 48
- Let the number be x.
- So, x = 48k + 3, where k is an integer.

Step 3: Check divisibility by 7
- x should be divisible by 7.
- Substitute the value of x from step 2 into the divisibility rule of 7.
- 48k + 3 should be divisible by 7.
- By trial and error, we find that when k = 3, 48(3) + 3 = 147 is divisible by 7.

Step 4: Verify the remainders
- When 147 is divided by 4, 12, and 16, it leaves a remainder of 3 in each case.
Therefore, the least number that satisfies all the given conditions is 147, which is option 'D'.
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Find the least number which when divided by 4, 12 and 16 leaves remainder 3 in each case, but it is perfectly divisible by 7.a)48b)51c)99d)147Correct answer is option 'D'. Can you explain this answer?
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