Q.Find all pairs of consecutive odd natural numbers, both of which are larger than 10, such that their sum is less than 40.
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Start your 14-day free trial to unlock the full solution →We need consecutive odd natural numbers > 10 whose sum < 40. Let the smaller be ; then the sum is , giving . With , we get . So , yielding the pairs (11,13), (13,15), (15,17), (17,19).
The problem asks for consecutive odd natural numbers — numbers like 1, 3, 5, 7, … where each is 2 more than the previous. Both must be larger than 10, and their sum must be less than 40.
The key is to translate these conditions into a simple inequality. Since the numbers are consecutive odds, if the smaller is , the next is . Both are natural numbers (positive integers), and we require and .
This is a linear inequality problem: we find the range of that satisfies both conditions, then list the odd numbers in that range.
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Set up the variables.
Let the smaller odd number be . Since they are consecutive odds, the larger is .
Both are natural numbers: .
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Apply the condition "both larger than 10".
The smaller must be > 10, so .
Since is odd, the smallest possible is 11.
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Apply the condition "sum is less than 40".
Sum: .
Subtract 2: .
Divide by 2: .
So must be an odd natural number satisfying .
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List the possible values.
Odd numbers between 10 and 19 (exclusive) are: 11, 13, 15, 17.
(Note: 19 is not allowed because , not .)
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Write the corresponding pairs. …
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