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Binomial Theorem question

2023 · 25 Jan · Shift 2 · Q41
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  5. /2023 · 25 Jan · Shift 2 · Q41

Binomial Theorem question

2023 · 25 Jan · Shift 2 · Q41

JEE MainMathematicsBinomial TheoremNumerical+4 / −1
The remainder when (2023) 2023^{2023}2023 is divided by 35 is ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 7

  1. We need the remainder when 202320232023^{2023}20232023 is divided by 353535.

  2. First reduce the base modulo 353535: 2023÷352023 \div 352023÷35 Since 35×57=1995,35 \times 57 = 1995,35×57=1995, we get 2023−1995=28.2023 - 1995 = 28.2023−1995=28. So, 2023≡28(mod35).2023 \equiv 28 \pmod{35}.2023≡28(mod35). Hence, 20232023≡282023(mod35).2023^{2023} \equiv 28^{2023} \pmod{35}.20232023≡282023(mod35).

  3. Now write 28≡−7(mod35).28 \equiv -7 \pmod{35}.28≡−7(mod35). So, 282023≡(−7)2023(mod35).28^{2023} \equiv (-7)^{2023} \pmod{35}.282023≡(−7)2023(mod35). But working directly with powers of −7-7−7 modulo 353535 is not convenient, so use the Chinese Remainder Theorem by considering mod 555 and mod 777.

  4. Modulo 555: 2023≡3(mod5).2023 \equiv 3 \pmod{5}.2023≡3(mod5). Therefore, 20232023≡32023(mod5).2023^{2023} \equiv 3^{2023} \pmod{5}.20232023≡32023(mod5). Now powers of 333 modulo 555 cycle with period 444:

\quad 3^2 \equiv 4, \quad 3^3 \equiv 2, \quad 3^4 \equiv 1 \pmod{5}.$$ Since $$2023 \equiv 3 \pmod{4},$$ we get $$3^{2023} \equiv 3^3 \equiv 27 \equiv 2 \pmod{5}.$$ So, $$2023^{2023} \equiv 2 \pmod{5}.$$ 5. Modulo $7$: $$2023 \equiv 0 \pmod{7}$$ because $2023 = 7 \times 289$. Thus, $$2023^{2023} \equiv 0^{2023} \equiv 0 \pmod{7}.$$ So, $$2023^{2023} \equiv 0 \pmod{7}.$$ 6. Now find the number $r$ such that $$r \equiv 2 \pmod{5},$$ $$r \equiv 0 \pmod{7}.$$ Multiples of $7$ are: $$0,7,14,21,28,\dots$$ Checking modulo $5$: - $0 \equiv 0 \pmod{5}$ - $7 \equiv 2 \pmod{5}$ So, $$r = 7.$$ 7. Therefore, the remainder when $2023^{2023}$ is divided by $35$ is $$\boxed{7}.$$
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