An algebra problem by Viki Zeta

Algebra Level pending

Let $A$ denote the set of an arithmetic progression (AP) $a_1, a_2, a_3, \ldots, a_{100}$ such that $a_n = 5 + (n-1)3$ and $B$ denote the set of an AP $b_1, b_2, b_3, \ldots, b_{100}$ such that $b_n = 3 + (n-1)4$ . Such that $n \ge 1$

Find how many common terms are there in both the sets $A$ and $B$ .

Clarifications:

'Common Terms' not necessarily mean that $a_n = b_n$ . It means for some value of items in $A$ and $B$ they are equal.

Hint :

Above clarification other words can be generalized as $a_n = b_m, ~ (a, b) \in \mathbb Z$ .

The answer is 25.

1 solution

Viki Zeta
Nov 2, 2016

$a_n = b_m \\ a + (n-1)d_1 = b + (m-1)d_2 \\ 5 + (n-1)3 = 3 + (m-1)4 \\ 2 + 3n - 3 = 4m - 4 \\ 3n - 1 = 4m - 4\\ 3n +3 = 4m\\ 3(n+1) = 4m\\ n = \dfrac{4m - 3}{3} = \dfrac{4m}{3} - 1 ~~ m = \dfrac{3n+3}{4} \\ \implies n < 101 \\ \implies \dfrac{4m}{3} - 1 < 101 \\ \implies \dfrac{4m}{3} < 102 \\ \implies 4m < 306 \\ \implies m < 76.5 \\ \implies m \le 76 \\ \text{also; } ~~ n \le 100\\ \boxed{\therefore \text{Equal no of terms = } n - m + 1 = 100 - 76 + 1= 25}$

The answer is 25. You're welcome to verify it yourself:

$(a_n)$ : (5, 8, 11 , 14, 17, 20, 23 , 26, 29, 32, 35 , 38, 41, 44, 47 , 50, 53, 56, 59 , 62, 65, 68, 71 , 74, 77, 80, 83 , 86, 89, 92, 95 , 98, 101, 104, 107 , 110, 113, 116, 119 , 122, 125, 128, 131 , 134, 137, 140, 143 , 146, 149, 152, 155 , 158, 161, 164, 167 , 170, 173, 176, 179 , 182, 185, 188, 191 , 194, 197, 200, 203 , 206, 209, 212, 215 , 218, 221, 224, 227 , 230, 233, 236, 239 , 242, 245, 248, 251 , 254, 257, 260, 263 , 266, 269, 272, 275 , 278, 281, 284, 287 , 290, 293, 296, 299 , 302)

$(b_n)$ : (3, 7, 11 , 15, 19, 23 , 27, 31, 35 , 39, 43, 47 , 51, 55, 59 , 63, 67, 71 , 75, 79, 83 , 87, 91, 95 , 99, 103, 107 , 111, 115, 119 , 123, 127, 131 , 135, 139, 143 , 147, 151, 155 , 159, 163, 167 , 171, 175, 179 , 183, 187, 191 , 195, 199, 203 , 207, 211, 215 , 219, 223, 227 , 231, 235, 239 , 243, 247, 251 , 255, 259, 263 , 267, 271, 275 , 279, 283, 287 , 291, 295, 299 , 303, 307, 311, 315, 319, 323, 327, 331, 335, 339, 343, 347, 351, 355, 359, 363, 367, 371, 375, 379, 383, 387, 391, 395, 399)

Ivan Koswara - 4 years, 7 months ago

Uh! Sorry. IDK but that reference said it was 24. And I made a mistake in my soln too. It's $n-m+1$ terms. I'll edit the question.