StatstriveISI intelligencePractice
Back to MSQE practice
PEAMCQModerate

2025 PEA Q20

Find 62025mod  256^{2025} \mod 25.

Reveal answer and solution

Answer

B

Solution

  1. 1

    Since gcd⁡(6,25)=1\gcd(6,25)=1, by Euler's theorem with φ(25)=20\varphi(25)=20,

  2. 2
    620≡1(mod25). 6^{20} \equiv 1 \pmod{25}.
  3. 3

    Write 2025=20⋅101+52025 = 20\cdot 101 + 5. Then

  4. 4
    62025≡65(mod25). 6^{2025} \equiv 6^{5} \pmod{25}.
  5. 5

    Compute: 62=36≡116^2 = 36 \equiv 11, 64≡112=121≡121−4⋅25=21≡−46^4 \equiv 11^2 = 121 \equiv 121 - 4\cdot 25 = 21 \equiv -4, 65≡−4⋅6=−24≡1(mod25)6^5 \equiv -4 \cdot 6 = -24 \equiv 1 \pmod{25}.

Answer structure / marking notes

Note 65≡1(mod25)6^5 \equiv 1 \pmod{25}, so 66 has order 55 mod 2525, and 20252025 is a multiple of 55.

%==========================================

Content note

Imported from public/resources/isi/msqe/solutions/pea/2025/ISI_MSQE_PEA_2025_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.