$8 + 88 + 888 + 8888 + 88888 + 888888 = x$

Concept & Strategy The problem asks for the sum of a series where the digit $8$ repeats and grows in length. The most efficient way to solve this is by factoring out the repeating digit or by vertically aligning the numbers and summing column by column from right to left. Step-by-Step Solution * Write the numbers vertically aligned by their unit digits: $$8$$ $$88$$ $$888$$ $$8888$$ $$88888$$ $$888888$$ * Sum the units column: There are six $8$s. $6 \times 8 = 48$. Write down $8$, carry over $4$. * Sum the tens column: There are five $8$s. $5 \times 8 = 40$. Add the carry ($40 + 4 = 44$). Write down $4$, carry over $4$. * Sum the hundreds column: There are four $8$s. $4 \times 8 = 32$. Add the carry ($32 + 4 = 36$). Write down $6$, carry over $3$. * Sum the thousands column: There are three $8$s. $3 \times 8 = 24$. Add the carry ($24 + 3 = 27$). Write down $7$, carry over $2$. * Sum the ten-thousands column: There are two $8$s. $2 \times 8 = 16$. Add the carry ($16 + 2 = 18$). Write down $8$, carry over $1$. * Sum the lakh column: There is one $8$. Add the carry ($8 + 1 = 9$). Write down $9$. * The final sum is $987648$. Exam Strategy & Shortcut You can often solve this using just the last few digits. Calculate the units place: $8 \times 6 = 48$ (ends in $8$). Calculate the tens place: $8 \times 5 + 4 = 44$ (ends in $48$). Look at the options: only (a), (b), (c), and (d) end in $48$. Calculate the hundreds place: $8 \times 4 + 4 = 36$ (ends in $648$). Only (a), (c), and (d) end in $648$. The thousands place: $8 \times 3 + 3 = 27$ (ends in $7648$). Only (a) and (d) remain. Ten-thousands place: $8 \times 2 + 2 = 18$ (ends in $87648$). This uniquely identifies option (d). Common Pitfall Attempting to add the numbers horizontally without proper alignment frequently leads to place value errors. Always stack them or use the digit counting method. Final Answer **Therefore, the correct answer is 987648.**