Directions : Solve the number series to answer the following questions. ⁿ√A 29…

Concept

In a number series, first decode every labelled placeholder from the stated conditions, then find the rule connecting consecutive terms. Two ideas are used here: (1) a number of the form p × q with two distinct primes p, q has exactly four divisors, namely {1, p, q, pq}; and (2) the successive differences in this series follow the closed form k³ + 1 for consecutive integers k.

Application

1. Find A and X.

   X is the difference between 29 and B. Excluding X itself, its factors are 13, 5 and A. Every integer greater than 1 has 1 among its factors, so the smallest of these three must be 1; hence A = 1. The factor list excluding X is then {1, 5, 13}, and including X gives 4 factors in all. The only number whose complete divisor set is {1, 5, 13, X} is 5 × 13 = 65, so X = 65.

2. Find B.

   |29 − B| = X = 65, and B must be a positive term, so B = 29 + 65 = 94.

3. Resolve the end terms using A.

   Since A = 1, the radical ⁿ√A = ⁿ√1 = 1 for every positive integer n. So the first term equals 1 and the last term 1294 / ⁿ√A = 1294 / 1 = 1294. The series so far is: 1, 29, 94, C, 437, (B + 687 = 781), 1294.

4. Identify the difference rule and find C.

   1. 1 → 29: difference 28 = 33 + 1

   2. 29 → 94: difference 65 = 43 + 1

   3. 94 → C: difference = 53 + 1 = 126, so C = 94 + 126 = 220

   4. 220 → 437: difference 217 = 63 + 1 (confirms 437)

   5. 437 → 781: difference 344 = 73 + 1 (confirms B + 687)

   6. 781 → 1294: difference 513 = 83 + 1 (confirms 1294 / ⁿ√A)

5. Compute A + C.

   A + C = 1 + 220 = 221.

Cross-check

Rebuilding the whole series from 1 with differences 3³+1, 4³+1, …, 8³+1 gives 1, 29, 94, 220, 437, 781, 1294 — every supplied term (29, 437, B + 687 = 781, and 1294 / ⁿ√A = 1294) matches, confirming C = 220 and A + C = 221.