Mathematics - Sequence & Series Question with Solution | TestHub
Match each item in List I to all applicable entries in List II.
List I | List II |
|---|---|
(A) Count divisors of N=2³·3²·5⁵·7⁴ leaving remainder 1 modulo 4. | (P) 16 |
(B) If a₁,...,a₁₀₀ form an H.P., find [Σ(i=1 to 99) aᵢaᵢ₊₁]/(a₁a₁₀₀). | (Q) 48 |
(C) Find 3³³ modulo 75. | (R) 126 |
(D) Count four-digit positive integers with each successive digit strictly larger than the preceding digit. | (S) 36 |
| (T) 99 |
Options:
Answer:
Solution:
(A) A divisor congruent to 1 modulo 4 has no factor 2. Exponents of 3 and 7 must have the same parity: (2·3+1·2) choices. There are six choices for the exponent of 5, giving 48. (B) If reciprocal A.P. difference d≠0, aᵢ-aᵢ₊₁=d aᵢaᵢ₊₁, and telescoping together with 1/a₁₀₀-1/a₁=99d yields the ratio 99. If d=0 all terms are equal, and the ratio is again 99. (C) 3³³ is 0 modulo 3 and 23 modulo 25, since 3²⁰≡1 and 3¹³≡23 modulo 25. The unique residue modulo 75 is 48. (D) Choose four distinct digits from 1,...,9; their increasing order is fixed. The count is C(9,4)=126. Thus A-Q; B-T; C-Q; D-R.
