AMC 8 Daily Practice - Multiplication Formula

Complete problem set with solutions and individual problem pages

Problem 4 Easy

If A = (2+1)(2^2+1)(2^4+1)\cdots(2^{32}+1)+1, what is the units digit of A?

  • A.

    0

  • B.

    2

  • C.

    4

  • D.

    6

  • E.

    8

Answer:D

A = (2+1)(2^2+1)(2^4+1)\dots(2^{32}+1)+1\\

=(2-1)(2+1)(2^2+1)(2^4+1)\dots(2^{32}+1)+1 \\

=(2^2-1)(2^2+1)(2^4+1)\dots(2^{32}+1)+1 \\

=(2^4-1)(2^4+1)\dots(2^{32}+1)+1 \\

=(2^{32}-1)(2^{32}+1)+1 \\

= 2^{64}-1+1 \\

= 2^{64}

Since 2^n cycles every 4: 2^1=2, 2^2=4, 2^3=8, 2^4=6, \dots

64 \div 4 = 16 remainder 0, so 2^{64} has units digit 6.

Final result: The units digit of A is \boxed{6}.

Related Problems
Problem 2 AMC 8 Daily Practice - Multiplication Formula
Problem 3 AMC 8 Daily Practice - Multiplication Formula
Problem 5 AMC 8 Daily Practice - Multiplication Formula
Problem 6 AMC 8 Daily Practice - Multiplication Formula
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