Recall that a composite number is a nonprime positive integer. We will call a composite number n a simple composite if it is the product of distinct primes and a complex composite if it has a factor of the form p² for some prime p.
       Write 12,167 in the form a₁ + a₂ + ××× + a₉ + b where each a_i is a complex composite and b is a simple composite.

Weng Shao

Observe that (x + y + z)³ = x³ + 3x²y + 3x²z + 3xy² + 6xyz + 3xz² + y³ + 3y²z + 3yz² + z³.
So
12,617 = (23)³ = (5 + 7+11)³ = 5³ + 3×5²×7 + 3×5² ×11 + 3 ×5 ×7² + 3 ×5 ×11² + 7³ + 3× 7² ×11 + 3 ×7×11² + 11³ + 3×2× 5× 7× 11

Nathan Collins

a₁ = a₂ =  ··· = a₉ = 2²  and b = 7 · 1733

36 + 12131 = 12167

Josiah Thornton 

23³ = (21 + 2)³ = 21³ + 3× 21² ×2 + 3 × 21× 2² + 2³ = 
21 ³ + 2906 = (19 + 2)³ + 2906 =

19³ + (3× 19² ×2 + 3 ×19× 2² + 2³ ) + 2906 = 17³ + 3× 17² ×2 + 3 ×17× 2² + 2³ + (3× 19² ×2 + 3 ×19× 2² + 2³ )
+ 2906 =  12167
a₁ = 17³, a₂ = a₃ = 3 ×19× 2, a₄ = a₅ =  3× 17², a₆ =  3 ×19× 2² , a₇ = 3 ×17× 2², a₈ = a₉ = 2³
b = 2906 = 2× 1453

Travis Willse

      The sum of the digits of 12167 is 17 = 8 (mod 9), so the largest multiples of 9 smaller than 12167 are 12159, 12150, 12141, 12132... The last two digits of 12132 are 32, a multiple of 4, so 12132 is a multiple of 4, and so is (12132)/9 = 1348. Set a_i = 1348 for i = 1, . . . , 9 (2² | 1348, so each a_i is a "complex composite") and b = 12167-12132 = 35 = 5 · 7 (b thus a simple composite). Then, we have 12167 written in the required form:
12167 = 1348 + · · · + 1348 + 35
       By adding and subtracting multiples of 4 from the a_i, we can write 12167 in the required form with distinct a_i, too:
12167 = 1332 + 1336 + · · · + 1360 + 1364 + 35