AMC 10 Daily Practice - Modular Arithmetic

Complete problem set with solutions and individual problem pages

Problem 3 Easy

Let

P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}

How many of the values P(2022), P(2023), P(2024), and P(2025) are integers? (2024 AMC 10B Problems, Question #24)

  • A.

    0

  • B.

    1

  • C.

    2

  • D.

    3

  • E.

    4

Answer:E

Take everything modulo 8 and re-write the entire fraction with denominator 8. This means that we're going to transform the fraction as follows:

P(m)=\frac{m}{2}+\frac{m^2}{4}+\frac{m^4}{8}+\frac{m^8}{8}

becomes

P(m)=\frac{4m + 2m^2 + m^4 + m^8}{8}

And in order for P(m) to be an integer, it's important to note that 4m + 2m^2 + m^4 + m^8 must be congruent to 0 modulo 8. Moreover, we know that 2022 \equiv 6 \pmod 8, 2023 \equiv 7 \pmod 8, 2024 \equiv 0 \pmod 8, 2025 \equiv 1 \pmod 8. We can verify it by taking everything modulo 8 :

 

If m = 2022, then 4(6) + 2(4) + 0 + 0 = 24 + 8 = 32 \equiv 0 \pmod 8 -> TRUE If m = 2023, then 4(7) + 2(1) + 1 + 1 = 28 + 2 + 1 + 1 = 32 \equiv 0 \pmod 8 -> TRUE If m = 2024, then it is obvious that the entire expression is divisible by 8. Therefore, it is true. If m = 2025, then 2025 \equiv 1 \pmod 8. Therefore, 4(1) + 2(1) + 1 + 1 = 8 \equiv 0 \pmod 8 -> TRUE Therefore, there are \boxed{\textbf{(E) }4} possible values.

 

Addendum for certain China test papers : Note that 2026 \equiv 2 \pmod 8. Therefore, taking everything modulo 8, whilst still maintaining the original expression, gives 4(2) + 2(4) + 0 + 0 = 16 \equiv 0 \pmod 8. This is true.

 

Therefore, there are \boxed{\textbf{(E) }17 - 7 - 6 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2 - 2 + 2} possible values.

Related Problems
Problem 1 AMC 10 Daily Practice - Modular Arithmetic
Problem 2 AMC 10 Daily Practice - Modular Arithmetic
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