Bunuel wrote:
OUT OF THE SCOPE - TOUGH QUESTION
A checkerboard of 13 rows and 17 columns has a number written in each square, beginning in the upper left corner, so that the first row is numbered 1, 2, ..., 17, the second row 18, 19, ..., 34, and so on down the board. If the board is renumbered so that the left column, top to bottom, is 1, 2, ...,13, the second column 14, 15, ..., 26 and so on across the board, some squares have the same numbers in both numbering systems. Find the sum of the numbers in these squares (under either system).
(A) 222
(B) 333
(C) 444
(D) 555
(E) 666
A lot of visualization is requiredSay we call the rows as r... 1, 2...r..13, and columns as c...1,2...c..16,17
So in the first case, what is 18 = 17*1+1, 20=17*1+3..
so what is 1 - it is the (row number-1)
What is 1, 3 and so on, it is the column number
Algebraic way to write about a particular box - 17(r-1)+c=17r-17+c
Similarly in the second case, what is 18 = 13*1+5, 20=13*1+7..
so what is 1 - it is the (column number-1)
What is 5, 7 and so on, it is the row number
Algebraic way to write about a particular box - 13(c-1)+r=13c-13+r
We are looking for cases when both are equal..
so 17r-17+c=13c-13+r.....16r=12c+4.....4r=3c+1
so 3c+1 will be a multiple of 4... It will happen when c=1, 5, (1+4k)
so cases are
when k=0, c=(1+4k)=1 and r =1, the number = 13c-13+r=13*1-13+1=1
when k=1, c=(1+4k)=5 and r =4, the number = 13c-13+r=13*5-13+4=56
when k=2, c=(1+4k)=9 and r =7, the number = 13c-13+r=13*9-13+7=111
The other 2 would be 166 and 221..
so sum = 1+56+111+166+221=555
D
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