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The sum of four consecutive odd integers in 2160. The greatest of them is
1. 543
2. 429
3. 537
4. 641

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Correct Answer - Option 1 : 543

Given:

The sum of four consecutive odd integers = 2160

Calculations:

Let the first integer be 'n'

The next three odd integers be 'n + 2', 'n + 4' , 'n + 6'

The sum number = n + n + 2 + n + 4 + n + 6

⇒ 4n + 12

4n + 12 = 2160

⇒ 4(n + 3) = 2160 

⇒ (n + 3) = 2160/4

⇒ n = 540 - 3 = 537

The greatest number n + 6 = 537 + 6 

⇒ 543

∴ The greatest of them is 543

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