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Today Puzzle #678

Puzzle No. 678– Tuesday 18 February

A bad tempered Maths teacher likes to give his students questions that they can’t use their calculators for. He asks in an exam: “What is the last digit of 7 to the power of 402?"

Today’s #PuzzleForToday has been set by Dr James Hind, is a Lecturer in Statistics in the School of Science and Technology at Nottingham Trent University

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You cannot do this on a calculator because the numbers become far too large. You can do it by trial and error, finding powers of seven until you spot a pattern or you can just think really hard!

1x7 is 7.

7x7 is 49.

The next time we multiply by Seven we will have 40 lots of Seven and Nine lots of Seven. The 40 lots of Seven must end in zero (280) because it must be a multiple of 40. We then add the remaining Nine lots of Seven: 9x7 is 63, so 3 is the last digit of 7x7x7.

This means to find the last digit of 7x7x7x7 we need 3x7=21 so 1 is the last digit. Next time we need 1x7 and that’s exactly where we started so we know what must come next.

The last digit of powers of Seven cycle 7, 9, 3, 1. For Seven to the power of 402 we have One Hundred full cycles and then get to the Second in the sequence, so the answer is 9.

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