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Binomial Theorem question

2023 · 13 Apr · Shift 1 · Q33
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  5. /2023 · 13 Apr · Shift 1 · Q33

Binomial Theorem question

2023 · 13 Apr · Shift 1 · Q33

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
Fractional part of the number 4202215\frac{4^{2022}}{15}1542022​ is equal to
  1. A
    815\frac{8}{15}158​
  2. B
    415\frac{4}{15}154​
  3. C
    115\frac{1}{15}151​
  4. D
    1415\frac{14}{15}1514​
View written solutionFree

Correct answer: C

  1. We need the fractional part of
4202215.\frac{4^{2022}}{15}.1542022​.

  1. The fractional part of N15\dfrac{N}{15}15N​ depends on the remainder when NNN is divided by 151515.

    If

    N=15q+r,0≤r<15,N = 15q + r, \quad 0 \le r < 15,N=15q+r,0≤r<15,

    then

    N15=q+r15,\frac{N}{15} = q + \frac{r}{15},15N​=q+15r​,

    so the fractional part is r15\dfrac{r}{15}15r​.

    Hence we need:

    42022(mod15).4^{2022} \pmod{15}.42022(mod15).

  2. Compute powers of 444 modulo 151515:

    41≡4(mod15),4^1 \equiv 4 \pmod{15},41≡4(mod15), 42=16≡1(mod15).4^2 = 16 \equiv 1 \pmod{15}.42=16≡1(mod15).

    Therefore powers repeat with period 222:

    42k≡1(mod15).4^{2k} \equiv 1 \pmod{15}.42k≡1(mod15).

  3. Since 202220222022 is even,

    42022=42⋅1011≡1(mod15).4^{2022} = 4^{2\cdot 1011} \equiv 1 \pmod{15}.42022=42⋅1011≡1(mod15).

    So the remainder on division by 151515 is 111.

  4. Therefore,

    4202215=integer+115.\frac{4^{2022}}{15} = \text{integer} + \frac{1}{15}.1542022​=integer+151​.

    Hence the fractional part is

    115.\boxed{\frac{1}{15}}.151​​.

  5. Option check:

    • A: 815\frac{8}{15}158​ — incorrect
    • B: 415\frac{4}{15}154​ — incorrect
    • C: 115\frac{1}{15}151​ — correct
    • D: 1415\frac{14}{15}1514​ — incorrect
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