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Worked Examples · Example 8

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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We need consecutive odd natural numbers > 10 whose sum < 40. Let the smaller be 2n+12n+1; then the sum is 4n+4<404n+4 < 40, giving n<9n < 9. With 2n+1>102n+1 > 10, we get n≥5n \ge 5. So n=5,6,7,8n = 5, 6, 7, 8, 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 xx, the next is x+2x+2. Both are natural numbers (positive integers), and we require x>10x > 10 and x+(x+2)<40x + (x+2) < 40.

This is a linear inequality problem: we find the range of xx that satisfies both conditions, then list the odd numbers in that range.


  1. Set up the variables.

    Let the smaller odd number be xx. Since they are consecutive odds, the larger is x+2x+2.

    Both are natural numbers: x∈{1,3,5,… }x \in \{1, 3, 5, \dots\}.

  2. Apply the condition "both larger than 10".

    The smaller must be > 10, so x>10x > 10.

    Since xx is odd, the smallest possible xx is 11.

  3. Apply the condition "sum is less than 40".

    Sum: x+(x+2)=2x+2<40x + (x+2) = 2x + 2 < 40.

    Subtract 2: 2x<382x < 38.

    Divide by 2: x<19x < 19.

    So xx must be an odd natural number satisfying 10<x<1910 < x < 19.

  4. List the possible values.

    Odd numbers between 10 and 19 (exclusive) are: 11, 13, 15, 17.

    (Note: 19 is not allowed because x<19x < 19, not ≤\le.)

  5. Write the corresponding pairs. …

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