Consider an arithmetic series and a geometric series having four initial terms from the set {11, 8, 21, 16, 26, 32, 4}. If the last terms of these series are the maximum possible four digit numbers, then the number of common terms in these two series is equal to ___________.

Solution

Maths sir · 4 steps

Pick out the AP and the GP from the set, then ask which powers of 2 leave remainder 1 on division by 5.

1. Split the set. The numbers 11, 16, 21, 26 form an AP with common difference 5. The numbers 4, 8, 16, 32 form a GP with common ratio 2. That uses up all seven numbers, with 16 shared by both.
2. Last terms. AP terms are 11 + 5k. The largest one up to 9999 is 11 + 5 × 1997 = 9996. GP terms are 2ⁿ for n ≥ 2. The largest four-digit power of 2 is 2¹³ = 8192.
3. Condition for common. Every AP term leaves remainder 1 when divided by 5. So a GP term 2ⁿ is common only if 2ⁿ ≡ 1 (mod 5), and it must also lie between 11 and 8192.

   n (mod 4)	Remainder of 2ⁿ on division by 5	In the AP?
   1	2	no
   2	4	no
   3	3	no
   0	1	yes

4. So. n must be a multiple of 4 with 2 ≤ n ≤ 13, so n = 4, 8, 12. The common terms are 16, 256, 4096, and the answer is 3.

The idea to keep. To find common terms of an AP and a GP, reduce the GP terms modulo the AP's common difference. The remainders of powers repeat in a short cycle, so you only have to check one cycle.

Answer from the NTA final answer key. Solution by PrepWiser.