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The smallest number which when diminished by 3 is divisible by 21, 28, 36 and 45 is

  • a)
    1163

  • b)
    1263

  • c)
    1283

  • d)
    1293

Correct answer is option 'B'. Can you explain this answer?
Verified Answer
The smallest number which when diminished by 3 is divisible by 21, 28,...
LCM of 21, 28, 36 and 45 is



∴ LCM of 21, 28, 36 and 45

= 3 × 2 × 3 × 2 × 7 × 5

= 36 × 35 

= 1260

∴ Required number = 1260 + 3 = 1263
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Most Upvoted Answer
The smallest number which when diminished by 3 is divisible by 21, 28,...
To find the smallest number that is divisible by 21, 28, 36, and 45, we need to find the least common multiple (LCM) of these numbers.

LCM is the smallest multiple that is divisible by all given numbers.

To find the LCM, we can use prime factorization and the highest power of each prime factor.

Prime factorization of the given numbers:
- 21 = 3 * 7
- 28 = 2^2 * 7
- 36 = 2^2 * 3^2
- 45 = 3^2 * 5

Now we need to choose the highest powers of the prime factors that appear in any of the given numbers.

- The highest power of 2 is 2^2
- The highest power of 3 is 3^2
- The highest power of 5 is 5
- The highest power of 7 is 7

To find the LCM, we multiply these highest powers together:
2^2 * 3^2 * 5 * 7 = 4 * 9 * 5 * 7 = 1260

Therefore, the smallest number that is divisible by 21, 28, 36, and 45 is 1260.

Since the question asks for the smallest number when divided by 3, we divide 1260 by 3:

1260 / 3 = 420

Therefore, the smallest number that is divisible by 21, 28, 36, and 45 when divided by 3 is 420.

However, none of the options given match the answer 420. Therefore, there may be an error in the options provided or in the given question.
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The smallest number which when diminished by 3 is divisible by 21, 28,...
Mam please find the answer
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