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A set of $15$ distinct positive integers $a_1,a_2,a_3,\dots,a_{15}$ is such that for each $i$, $2\le i\le 15$, $a_i=a_{i-1}+i,a_i=a_{i-1}-i,a_i=a_{i-1}\times i$, or $a_i=a_{i-1}\div i$. If the largest of these $15$ integers is $19$, find all the others.

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  • $\begingroup$ Ten such distinct positive integers, the largest of which is $13$, are $1, 2, 6, 10, 5, 11, 4, 12, 3, 13$. $\endgroup$ Commented Jan 3, 2017 at 3:41

1 Answer 1

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Solutions:

1,+2,3,×3,9,+4,13,-5,8,+6,14,÷7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
9,+2,11,+3,14,-4,10,÷5,2,+6,8,+7,15,-8,7,+9,16,-10,6,+11,17,-12,5,+13,18,-14,4,+15,19
10,-2,8,+3,11,+4,15,÷5,3,+6,9,-7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
10,+2,12,-3,9,+4,13,-5,8,+6,14,÷7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
11,-2,9,+3,12,÷4,3,+5,8,+6,14,÷7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
11,+2,13,-3,10,+4,14,-5,9,+6,15,-7,8,+8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
12,-2,10,+3,13,-4,9,+5,14,-6,8,+7,15,-8,7,+9,16,-10,6,+11,17,-12,5,+13,18,-14,4,+15,19
12,+2,14,-3,11,+4,15,÷5,3,+6,9,-7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
13,-2,11,+3,14,-4,10,÷5,2,+6,8,+7,15,-8,7,+9,16,-10,6,+11,17,-12,5,+13,18,-14,4,+15,19
13,-2,11,+3,14,-4,10,+5,15,-6,9,-7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
13,+2,15,-3,12,÷4,3,+5,8,+6,14,÷7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4
13,+2,15,-3,12,-4,8,-5,3,+6,9,-7,2,×8,16,-9,7,+10,17,-11,6,+12,18,-13,5,+14,19,-15,4

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  • $\begingroup$ The intended solution is the first above in which the first term of the sequence is 1. $\endgroup$ Commented Jan 3, 2017 at 3:58
  • 2
    $\begingroup$ Ah. That wasn't specified in the problem, so I provided all of them :) $\endgroup$
    – Rubio
    Commented Jan 3, 2017 at 4:05
  • $\begingroup$ So much the better! $\endgroup$ Commented Jan 3, 2017 at 4:06

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