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The sequence 12,15, 18, 21, 51, 81,... consists of all positive multiples of 3 that contain at least one digit that is a 1. what is the 80th term of this sequence?

 Jun 23, 2021
 #1
avatar+20 
+2

Your sequence is:

 

(12, 15, 18, 21, 51, 81, 102, 105, 108, 120, 123, 126, 129, 132, 135, 138, 144, 147, 150, 153, 156, 159, 162, 165, 168, 174, 177, 180, 183, 186, 189, 192, 195, 198, 201, 210, 213, 216, 219, 231, 261, 291, 312, 315, 318, 321, 351, 381, 414, 417, 441, 471, 501, 510, 513, 516, 519, 531, 561, 591, 612, 615, 618, 621, 651, 681, 714, 717, 741, 771, 801, 810, 813, 816, 819, 831, 861, 891, 912, 915)>>Total =80 terms

 

The 80th term = 915

 Jun 23, 2021
 #2
avatar+287 
0

Excuse me for not understanding.  If the sequence consists of all positive multiples of 3 that contain at least one digit that is a 1, shouldn't numbers like 111, 114, 141, 117, 171, etc. belong to the sequence as well?
 

Bginner  Jun 24, 2021

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