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Problem #51 · September 16, 2026

Last digit

Find the last digit of the sum 7+72+73+⋯+7507 + 7^{2} + 7^{3} + \cdots + 7^{50}.
  1. A0
  2. B6
  3. C4
  4. D7

WORKED EXAMPLE

Solution

The last digits of the terms repeat in the cycle 7,9,3,17, 9, 3, 1, and summing each group of 4 such digits gives 7+9+3+1=207+9+3+1=20, ending in 00. 50=4⋅12+250 = 4 \cdot 12 + 2: the first 4848 terms form 1212 complete cycles (ending in 00), and the remaining two terms 7497^{49} and 7507^{50} end in 77 and 99, respectively. Total: 0+7+9=160 + 7 + 9 = 16, with last digit 66.
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