Arrange the numbers from 1 to 20 in a row such that the sum of any two adjacent numbers is a perfect square
Answers
Answer:
Hope this helps.
Step-by-step explanation:
For 20 it's not actually possible
But it can happen for 15,17,25
15 - > [8,1,15,10,6,3,13,12,4,5,11,14,2,7,9]
17 -> [16, 9, 7, 2, 14, 11, 5, 12, 13, 3, 6, 10, 15, 1, 8, 17]
20 -> No Solution
[ 11,14, 2, 7, 9, 16, 20, 5, 4, 12, 13, 3, 6, 19, 17, 8, 1, 15, 10
18] PROBLEM IS WITH 18
25 -> [2, 23, 13, 12, 24, 25, 11, 14, 22, 3, 1, 8, 17, 19, 6, 10,
15, 21, 4, 5, 20, 16, 9, 7, 18]
one can arrange all the numbers from 1 to 305 in a row such that the sum of every two adjacent numbers is a perfect cube.
256,87,129,214,298,45,171,172,44,299,213,130,86,257,255,
88,128,215,297,46,170,173,43,300,212,131,85,258,254,89,127,216,296,
47,169,174,42,301,211,132,84,259,253,90,126,217,295,48,168,175,41,302,
210,133,83,260,252,91,125,218,294,49,167,176,40,303,209,134,82,261,251,
92,33,183,160,56,287,225,118,98,245,267,76,140,203,13,14,202,141,75,268,
244,99,26,190,153,63,280,232,111,105,238,274,69,147,196,20,7,1,124,219,
293,50,166,177,39,304,208,135,81,262,250,93,32,184,159,57,286,226,117,8,
19,197,146,70,273,239,104,112,231,281,62,154,189,27,37,179,164,52,291,221,
122,3,5,22,194,149,67,276,236,107,109,234,278,65,151,192,24,101,242,270,
73,143,200,16,11,205,138,78,265,247,96,120,223,289,54,162,181,35,29,187,
156,60,283,229,114,102,241,271,72,144,199,17,108,235,277,66,150,193,23,
4,121,222,290,53,163,180,36,28,188,155,61,282,230,113,103,240,272,71,145,
198,18,9,116,227,285,58,158,185,31,94,249,263,80,136,207,305,38,178,165,
51,292,220,123,2,6,21,195,148,68,275,237,106,110,233,279,64,152,191,25,
100,243,269,74,142,201,15,12,204,139,77,266,246,97,119,224,288,55,161,
182,34,30,186,157,59,284,228,115,10,206,137,79,264,248,95