To find the values of c and k such that the number 3c2933k is divisible by both 5 and 11: Divisibility by 5: The last digit k must be 0 or 5. Divisibility by 11: The alternating sum and difference of the digits must be divisible by 11. Case-1, k = 0 Alternating sum S=3-c+2-9+3-3+0=-4-c For divisibility by 11=-4-c=0 (mod 11) c=7 Case-2: k=5 Alternating sum: S=3-c+2-9+3-3+5=5-c-4 For divisibility by 11=1-c=0 (mod 11) c=1 Final Solution: For k = 0 and c = 7, S = 3 (not divisible by 11). For k = 5 and c = 1, S = 0 (divisible by 11). Thus, the correct combination is c = 1 and k = 5. then now c+k =1+5 =6.
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