Let k = 147, k = 169, k3 = 580, k4 = 216, k5= 974, k6...

  

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Let k₁ = 147, k₂ = 169, k3 = 580, k4 = 216, k5= 974, k6 = 124. Suppose HT is of size 13 indexed 0,1,2,...,12. Define the function h: {k₁, k2, k3, k4, k5, k6} {0,1,2,...,12 } by → h(k) = k₁ %13. Now h(k)=h(147): = 147 % 13 = 4. So the data of the student with ID 147 is stored in HT[4]. We construct the following table that shows the array position where each student's data is stored. h(ID) (h(ID) + 1) % 13 ID 147 169 580 216 974 124 4 0 8 8 12 7 9 (h(ID) +2) % 13 Let k₁ = 147, k₂ = 169, k3 = 580, k4 = 216, k5= 974, k6 = 124. Suppose HT is of size 13 indexed 0,1,2,...,12. Define the function h: {k₁, k2, k3, k4, k5, k6} {0,1,2,...,12 } by → h(k) = k₁ %13. Now h(k)=h(147): = 147 % 13 = 4. So the data of the student with ID 147 is stored in HT[4]. We construct the following table that shows the array position where each student's data is stored. h(ID) (h(ID) + 1) % 13 ID 147 169 580 216 974 124 4 0 8 8 12 7 9 (h(ID) +2) % 13 Let k₁ = 147, k₂ = 169, k3 = 580, k4 = 216, k5= 974, k6 = 124. Suppose HT is of size 13 indexed 0,1,2,...,12. Define the function h: {k₁, k2, k3, k4, k5, k6} {0,1,2,...,12 } by → h(k) = k₁ %13. Now h(k)=h(147): = 147 % 13 = 4. So the data of the student with ID 147 is stored in HT[4]. We construct the following table that shows the array position where each student's data is stored. h(ID) (h(ID) + 1) % 13 ID 147 169 580 216 974 124 4 0 8 8 12 7 9 (h(ID) +2) % 13 Let k₁ = 147, k₂ = 169, k3 = 580, k4 = 216, k5= 974, k6 = 124. Suppose HT is of size 13 indexed 0,1,2,...,12. Define the function h: {k₁, k2, k3, k4, k5, k6} {0,1,2,...,12 } by → h(k) = k₁ %13. Now h(k)=h(147): = 147 % 13 = 4. So the data of the student with ID 147 is stored in HT[4]. We construct the following table that shows the array position where each student's data is stored. h(ID) (h(ID) + 1) % 13 ID 147 169 580 216 974 124 4 0 8 8 12 7 9 (h(ID) +2) % 13

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