2 Answers

Mirek Michalak · Answer 1 · 2021-01-05T15:30:32+0000

Rewrite the product as

$13^4\cdot 17^2\cdot29^3 = (10+3)^4 (10+7)^2 (20+9)^3$

Powers of 10 and multiples of 10 don't contribute to the units digits, so we only need to consider the product

$3^4\cdot7^2\cdot9^3=81\cdot49\cdot729$

and we can treat this the same way:

$81\cdot49\cdot729=(80+1)(40+9)(720+9)$

Again, ignore the multiples of 10 to reduce this to the product 1*9*9 = 81, whose units digit is 1.

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