A traveller arrived to an island where 50 natives lived. All the natives stood in a circle and each announced firstly the age of his left neighbour, then the age of his right neighbour. Each native is either a knight who told both numbers correctly or a knave who increased one of the numbers by 1 and decreased the other by 1 (on his choice). Is it always possible after that to establish which of the natives are knights and which are knaves?

since proposition states that a number is greater than 25, if is true then the number must actually be greater than 25.

this implies that all possible options, except 24, can be accepted

< xcy ≅ < xby

step-by-step explanation:

in δcax and δbax,

so, δcax ≅ δbax (side-angle-side or sas)

since δcax ≅ δbax, we can conclude < acx ≅ < abx

in δacy and δaby,

so, δacy ≅ δaby (side-angle-side or sas)

since δacy ≅ δaby, we can conclude < acy ≅ < aby

now,

< acy ≅ < aby

=> < acx + < xcy ≅ < abx + < xby

=> < acx + < xcy ≅ < acx + < xby   (since < acx ≅ < abx already proved above)

subtracting < acx from both sides, we get

< acx + < xcy -< acx ≅ < acx + < xby -< acx

cancelling out < acx and -< acx from both sides, we get

< xcy ≅ < xby (proved)

d 600 because 15*10*2=300

300+300=600

step-by-step explanation: