A number theory problem by Puneet Pinku

What is the largest even integer that cannot be written as the sum of two odd composite positive integers?


The answer is 38.

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1 solution

Zee Ell
Aug 31, 2016

Firstly, we will show that 38 cannot be written as a sum of two composite odd numbers.

For this, it is enough to show that the bigger (or one of the equal numbers, 19 in this case, which means that both are primes) number b cannot be composite.

Since s is the smaller number (when b is the bigger number), therefore

s + s s + b = 38 2 s 38 s 19 s + s ≤ s + b = 38 \iff 2s ≤ 38 \iff s ≤ 19

The composite odd numbers in the 0 < s ≤ 19 interval are:

  • 9 : 9 + 29 = 38 , 29 is a prime, therefore not a composite number

  • 15 : 15 + 23 = 38 , 23 is a prime, therefore not a composite number.

Hence, 38 cannot be written as a sum of two composite positive odd numbers.

Now, we will show that all N > 38 (even) numbers can be written as a sum of two odd positive integers.

Let's have a look at the following cases:

a) N = 3 k , k Z + , k 13 : N = 3k , k \in \mathbb {Z}^+ , k ≥ 13 :

With the choice of s = 9 (composite: 9 = 3 × 3; the smallest positive odd composite integer (divisible by 3)), we have

N = s + b = 3k = 9 + (3k - 9) = 9 + 3(k - 3)

b = 3(k - 3) , which is composite, as it has at least two factors:

3 and (k - 3) ≥ 10 > 1.

b) N = 3 k + 1 , k Z + , k 13 : N = 3k + 1 , k \in \mathbb {Z}^+ , k ≥ 13 :

With the choice of n 1 = 25 (composite: 25 = 5 × 5; \text {With the choice of } n_1 = 25 \text { (composite: 25 = 5 × 5; }

25 is the smallest positive odd composite integer (giving a remainder of 1 when divided by 3)), we have:

N = n 1 + n 2 = 3 k + 1 = 25 + ( 3 k 24 ) = 6 + 3 ( k 8 ) N = n_1 + n_2 = 3k +1 = 25 + (3k - 24) = 6 + 3(k - 8)

n 2 = 3 ( k 8 ) , which is composite, too, as it has at least two factors: n_2 = 3(k - 8) \text { , which is composite, too, as it has at least two factors: }

3 and (k - 8) ≥ 5 > 1.

c) N = 3 k + 2 , k Z + , k 13 : N = 3k + 2 , k \in \mathbb {Z}^+ , k ≥ 13 :

With the choice of n 1 = 35 (composite: 35 = 5 × 7 ; \text {With the choice of } n_1 = 35 \text { (composite: 35 = 5 × 7 ; }

35 is the smallest positive odd composite integer (giving a remainder of 2 when divided by 3)), we have: } )

N = n 1 + n 2 = 3 k + 2 = 35 + ( 3 k 33 ) = 6 + 3 ( k 11 ) N = n_1 + n_2 = 3k + 2 = 35 + (3k - 33) = 6 + 3(k - 11)

n 2 = 3 ( k 11 ) , which is composite, too, as it has at least two factors: n_2 = 3(k - 11) \text { , which is composite, too, as it has at least two factors: }

3 and (k - 11) ≥ 2 > 1.

Therefore, the largest even integer that cannot be written as the sum of two odd composite positive integers is:

38 \boxed {38}

Thanks for the solution!! Could you also solve this question?? https://brilliant.org/discussions/thread/algebra-mania/

Puneet Pinku - 4 years, 9 months ago

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Solved. See at your question.

Zee Ell - 4 years, 9 months ago

This is from AIME 1984 Problem 14 isn't it?

Chaebum Sheen - 4 years, 9 months ago

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I don't know, I found this question in one of the Indian exams!

Puneet Pinku - 4 years, 8 months ago

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A lot of nations (and many books) do copy US's math Olympiads without referencing them. :(

Chaebum Sheen - 4 years, 8 months ago

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