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Binomial Theorem question

2023 · 6 Apr · Shift 2 · Q29
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  5. /2023 · 6 Apr · Shift 2 · Q29

Binomial Theorem question

2023 · 6 Apr · Shift 2 · Q29

JEE MainMathematicsBinomial TheoremMCQ+4 / −1
Among the statements : (S1) : 20232022−199920222023^{2022}-1999^{2022}20232022−19992022 is divisible by 8 (S2) : 13(13)n−12n−1313(13)^{n}-12 n-1313(13)n−12n−13 is divisible by 144 for infinitely many n∈Nn \in \mathbb{N}n∈N
  1. A
    both (S1) and (S2) are incorrect
  2. B
    only (S1) is correct
  3. C
    only (S2) is correct
  4. D
    both (S1) and (S2) are correct
View written solutionFree

Correct answer: D

  1. Check statement (S1):

We need to test whether 20232022−199920222023^{2022}-1999^{2022}20232022−19992022 is divisible by 888.

Reduce the bases modulo 888: 2023≡7≡−1(mod8)2023\equiv 7 \equiv -1 \pmod{8}2023≡7≡−1(mod8) 1999≡7≡−1(mod8)1999\equiv 7 \equiv -1 \pmod{8}1999≡7≡−1(mod8)

Hence, 20232022≡(−1)2022=1(mod8)2023^{2022}\equiv (-1)^{2022}=1 \pmod{8}20232022≡(−1)2022=1(mod8) 19992022≡(−1)2022=1(mod8)1999^{2022}\equiv (-1)^{2022}=1 \pmod{8}19992022≡(−1)2022=1(mod8)

Therefore, 20232022−19992022≡1−1=0(mod8)2023^{2022}-1999^{2022}\equiv 1-1=0 \pmod{8}20232022−19992022≡1−1=0(mod8)

So (S1) is correct.

  1. Check statement (S2):

The expression is 13(13)n−12n−13=13n+1−12n−1313(13)^n-12n-13 = 13^{n+1}-12n-1313(13)n−12n−13=13n+1−12n−13

We need to see whether it is divisible by 144=12⋅12=16⋅9144=12\cdot 12=16\cdot 9144=12⋅12=16⋅9? Actually, 144=16×9144=16\times 9144=16×9 with gcd⁡(16,9)=1\gcd(16,9)=1gcd(16,9)=1. So we check divisibility by 161616 and 999.

Step 2.1: Check modulo 161616

We need 13n+1−12n−13≡0(mod16)13^{n+1}-12n-13 \equiv 0 \pmod{16}13n+1−12n−13≡0(mod16)

Now, 13≡−3(mod16)13\equiv -3 \pmod{16}13≡−3(mod16) Also powers of 131313 modulo 161616 cycle:

\quad 13^2\equiv 9, \quad 13^3\equiv 5, \quad 13^4\equiv 1 \pmod{16}$$ So period is $4$. Let us try $n\equiv 0 \pmod{4}$, say $n=4k$. Then $$13^{n+1}=13^{4k+1}=(13^4)^k\cdot 13 \equiv 1^k\cdot 13=13 \pmod{16}$$ And $$12n+13 = 48k+13 \equiv 13 \pmod{16}$$ Thus, $$13^{n+1}-12n-13 \equiv 13-13=0 \pmod{16}$$ So whenever $n\equiv 0\pmod{4}$, divisibility by $16$ holds. --- ### Step 2.2: Check modulo $9$ We need $$13^{n+1}-12n-13 \equiv 0 \pmod{9}$$ Reduce terms modulo $9$: $$13\equiv 4 \pmod{9}, \quad 12n\equiv 3n \pmod{9}$$ So condition becomes $$4^{n+1}-3n-4 \equiv 0 \pmod{9}$$ Now powers of $4$ modulo $9$ cycle with period $3$: $$4^1\equiv 4, \quad 4^2\equiv 7, \quad 4^3\equiv 1 \pmod{9}$$ Take again $n=4k$. Then modulo $3$, $4k\equiv k$, so choose in particular $n$ to be a multiple of $12$, i.e. $n=12m$. Then $$n+1=12m+1 \implies 4^{n+1}=4^{12m+1}=(4^3)^{4m}\cdot 4 \equiv 1\cdot 4=4 \pmod{9}$$ Also, $$3n+4 = 36m+4 \equiv 4 \pmod{9}$$ Hence, $$4^{n+1}-3n-4 \equiv 4-4=0 \pmod{9}$$ So whenever $n\equiv 0\pmod{12}$, divisibility by $9$ holds. --- ### Step 2.3: Combine both conditions If $$n\equiv 0 \pmod{12}$$ then certainly $n\equiv 0\pmod{4}$ also, so the expression is divisible by both $16$ and $9$. Therefore, for every $n=12m$, $$13^{n+1}-12n-13$$ is divisible by $144$. Since there are infinitely many multiples of $12$, **(S2) is correct for infinitely many $n\in\mathbb N$**. So **(S2) is correct**. --- 3. **Final conclusion** - (S1) is correct - (S2) is correct Hence the correct option is $$\boxed{\text{D}}$$ --- 4. **Comparison with stored answer** Stored correct answer: $\text{D}$ My derived answer is also $\text{D}$, so they agree.
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