Saturday, December
14, 2002 

NUMBERS are written in all 16 squares of a 4 by 4 table, so that, the sum of the neighbours of each number is 1. Neighbouring squares are those with a common side. Find the sum of all the numbers in the table. The sum of the following
six slots (marked by X) is the sum of all 16 numbers in the table. The
sum of the neighbour of each number is 1; and the sum is equal to 6: 
X 0 0 X 0 0 0 0 0 X X 0 "The answer is 4," says Gagandeep Kaur Atwal; she misses it by two points. "The sum of all numbers is 8," says Puneet Goyal of Nabha. Bhavesh Valecha of Naya Bazar in Sunam says that the sum is 7. "Let all 4 rows be numbered 1,2,3,4 and columns 1,2,3,4, such that each element's number is (column, row). Neighbours of cell (2,2): (1,2) + (2,3) + (2,1) + (3,2)=1; Neighbours of cell(3,2): (2,2)+(3,1)+(3,3)+(4,2)=1; Neighbours of cell(1,4): (1,3)+(2,4)=1; Neighbours of cell(4,4): (3,4)+(4,3)=1; Remaining cells: (1,1);(1,4);(4,1);(4,4); (1,1)+(2,2)+(1,3)=1 { NOW (2,2)+(1,3)+(2,4)+(3,3)=1; (2,2)+(1,3)=1[(2,4)+(3,3)]} ; (1,1)+(2,2)+1[(2,4)+(3,3)]} =1; (1,1)+(2,2)+1[1(4,4)]=1; (1,1)+(4,4)=1; Similarly, (1,4)+(4,1)=1, so, sum =6," says Bhupender Singh of Patiala "Let the grid be represented by letters A to P; nb=neibhour; nb of A=B+E=1; =>nb of F=B+E+G+J=1 =>G+J=0; similarly, C+H =O+L= N+I =B+E = 1 and G+J=F+K =0; G, F, J, K nb 4 no.s; B, E, I, N, O, L ,C, Hnb 3 no.s; A, D, M, P nb 2 no.s; adding nb of all 16 no.s: 4(G+J+F+K)+3(C+H+O+L+N+I+B+E)+2(A+D+M+P)=16 => A+D+M+P=2; therefore, G+J+F+K+C+H+O+L+N+I+B+E+G+J+F+K=6," says Deepankar Garg of Class XII, DAV Amritsar. "Numbers 1 to 16 denote the 16 unknown no.s and do not represent their actual values. The no.s on the extreme corners (Group1: 1, 4, 13, 16) have 2 neighbours each, while those sharing a single side with the extreme boundary (Group2: 2, 3, 5, 8, 9, 12, 14, 15) have 3 neighbours each; and those which don't share any side with the boundary (Group3: 6, 7, 10, 11) have 4 neighbours each. Divide these no.s into sets, such that, the sets are disjoint and many (or even all) of which can be expressed as neighbours of somenumber; and their sum can be written as one. Neighbours of no.s in group 1: too many no.s (8) are left; and group 3: problem in getting disjoint sets, so, we take group 2. Let S(T) denote the required sum and s(n,i) denote the sum of neighbors of i'th number. We have, S(T) = 2 + s(n,2) + 8 + s(n,8) + 15 + s(n,15) + 9 + s(n,9); Also, S(T) = 3 + s(n,3) + 12 + s(n,12) + 14 + s(n,14) + 5 + s(n,5) (all s(n,i)'s are disjoint). Since, each s(n,i) = 1, we can find another expression for S(T) in terms of s(n,i)'s and the remaining 8 no.s (1, 4, 6, 7, 10, 11, 13, 16), which, when added to the two equations and combined with the other 8 no.s, can be written as S(T). That expression is, S(T) = 1 + s(n,1) + 4 + s(n,4) + 16 + s(n,16) + 13 + s(n,13) + 6 + 7 +10 + 11. Adding all the three expressions for S(n) we have, 3[ S(T) ] = s(n,1) + s(n,2) + s(n,3) + s(n,4) + s(n,5) + s(n,8) + s(n,9) + s(n,12) + s(n,13) + s(n,14) +s(n,15) + s(n,16) + [ S(T) ]; 2 [ S(n) ] = 12 [ s (n,i) ]; 2 [ S(n) ] = 12 [ 1 ] = 12; [ S(n)] = 6," says Sushane Chopra, who is in Class Xll. Dr Tarsem Lal of Khanna,
Baljot Kaur, Deepinder Singh of Patiala, Sabita, Rohit Pardasani, Suhail
Singh Shergill Navdeep Singh and Varun Bhardwaj of PEC also strike it
rich with number 6. 