The product of the last two digits of (1919)¹⁹¹⁹$(1919)^{1919}$ is

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Core Logic

The general term in the expansion is:

For the term to be integral, both powers (n-r)/(3)$\frac{n-r}{3}$ and (r)/(12)$\frac{r}{12}$ must be integers. This dictates that r$r$ must be a multiple of 12$12$ (r = 12k$r = 12k$, where k$k$ is a non-negative integer).

Step 1: Setup total count equation

The values r$r$ can take are 0, 12, 24, , 12k$0, 12, 24, \dots, 12k$. The total number of terms is given as 183$183$. Since counting starts from k=0$k=0$, the maximum value of k$k$ is:

Step 2: Calculate minimum n

The maximum index value r$r$ required to achieve this count is:

Hence, the least value of n$n$ must be 2184$2184$ to encompass all 183 integral terms.

Pattern Recognition

Number of terms formula when tracking steps of size L$L$: Total Terms = (n)/(L) + 1$\text{Total Terms} = \lfloor \frac{n}{L} \rfloor + 1$. Isolate n$n$ directly to compute bounds rapidly.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

General term in a binomial expansion (1+t)ⁿ$(1+t)^n$ is given by:

Core Logic

Let's simplify the algebraic structure of the product expression first:

Finding the coefficient of x³⁰$x^{30}$ in this full product is equivalent to finding the coefficient of x³⁶$x^{36}$ in the numerator expansion:

Step 1: Setting up General Term Constraints

The product of the three general terms is:

We require the total exponent to equal 36$36$:

with boundaries 0 ≤ r₁ ≤ 6$0 \le r_1 \le 6$, 0 ≤ r₂ ≤ 7$0 \le r_2 \le 7$, 0 ≤ r₃ ≤ 8$0 \le r_3 \le 8$.

Step 2: Case Analysis by r3

Let's evaluate non-vanishing integer combinations case-by-case:

• Case I: r₃ = 8$r_3 = 8$

• r₂ = 6, r₁ = 0 60 76 88(-1)⁸ = 1 × 7 × 1 = 7$r_2 = 6, r_1 = 0 \implies \binom{6}{0}\binom{7}{6}\binom{8}{8}(-1)^8 = 1 \times 7 \times 1 = 7$
• r₂ = 5, r₁ = 2 62 75 88(-1)⁸ = 15 × 21 × 1 = 315$r_2 = 5, r_1 = 2 \implies \binom{6}{2}\binom{7}{5}\binom{8}{8}(-1)^8 = 15 \times 21 \times 1 = 315$
• r₂ = 4, r₁ = 4 64 74 88(-1)⁸ = 15 × 35 × 1 = 525$r_2 = 4, r_1 = 4 \implies \binom{6}{4}\binom{7}{4}\binom{8}{8}(-1)^8 = 15 \times 35 \times 1 = 525$
• r₂ = 3, r₁ = 6 66 73 88(-1)⁸ = 1 × 35 × 1 = 35$r_2 = 3, r_1 = 6 \implies \binom{6}{6}\binom{7}{3}\binom{8}{8}(-1)^8 = 1 \times 35 \times 1 = 35$
• Case II: r₃ = 7$r_3 = 7$

• r₂ = 7, r₁ = 1 61 77 87(-1)⁷ = 6 × 1 × 8 × (-1) = -48$r_2 = 7, r_1 = 1 \implies \binom{6}{1}\binom{7}{7}\binom{8}{7}(-1)^7 = 6 \times 1 \times 8 \times (-1) = -48$
• r₂ = 6, r₁ = 3 63 76 87(-1)⁷ = 20 × 7 × 8 × (-1) = -1120$r_2 = 6, r_1 = 3 \implies \binom{6}{3}\binom{7}{6}\binom{8}{7}(-1)^7 = 20 \times 7 \times 8 \times (-1) = -1120$
• r₂ = 5, r₁ = 5 65 75 87(-1)⁷ = 6 × 21 × 8 × (-1) = -1008$r_2 = 5, r_1 = 5 \implies \binom{6}{5}\binom{7}{5}\binom{8}{7}(-1)^7 = 6 \times 21 \times 8 \times (-1) = -1008$
• Case III: r₃ = 6$r_3 = 6$

• r₂ = 7, r₁ = 4 64 77 86(-1)⁶ = 15 × 1 × 28 = 420$r_2 = 7, r_1 = 4 \implies \binom{6}{4}\binom{7}{7}\binom{8}{6}(-1)^6 = 15 \times 1 \times 28 = 420$
• r₂ = 6, r₁ = 6 66 76 86(-1)⁶ = 1 × 7 × 28 = 196$r_2 = 6, r_1 = 6 \implies \binom{6}{6}\binom{7}{6}\binom{8}{6}(-1)^6 = 1 \times 7 \times 28 = 196$

Step 3: Summation for Alpha

Summing all calculated values:

Thus, the absolute value is:

|α| = 678$|\alpha| = 678$

Pattern Recognition

Sees: Multi-product polynomial coefficient extraction problem. Trap: Remember that the negative sign inside (1-x³)⁸$(1-x^3)^8$ alters the polarity of terms based on whether r₃$r_3$ is odd or even. Always track the (-1)r₃$(-1)^{r_3}$ factor carefully.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

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Core Logic

Let the giant power tower expression be variable parameter t = 32³²$t = 32^{32}$. Re-expressing baseline number base 64: 64 = 8²$64 = 8^2$ So, the total term evaluates to:

Step 1: Evaluating Modulo Properties

We can re-express baseline 8 as (9 - 1)$(9 - 1)$:

Expanding this expression using the Binomial Theorem:

Since the exponent 2t$2t$ is an even number, (-1)2t = 1$(-1)^{2t} = 1$. Therefore:

Thus, the remainder obtained when divided by 9 is exactly 1.

Pattern Recognition

When evaluating exponents modulo M$M$, reduce the base first. Since 64 ≡ 1 9$64 \equiv 1 \pmod 9$, any integer power of 64 trivially leaves a remainder of 1.

Chapter Mix

Class 11 Mathematics: Binomial Theorem

Related Formula

Core Logic

From the given equation:

Applying the combination property, we get:

Since n, r ≥ 0$n, r \ge 0$ and n ≥ r+1$n \ge r+1$ for the combination to be valid, the ratio (r+1)/(n)$\frac{r+1}{n}$ must satisfy:

Substituting this bound into our expression:

Step 1: Solving the Inequalities

First inequality:

Second inequality:

Taking the intersection of both intervals:

Step 2: Final Conclusion

Looking at the options, we consider the positive domain interval:

Pattern Recognition

Combinatorics identities often reduce to bounds on variables. Memorize the ratio ⁿ⁻¹CᵣⁿCᵣ₊₁ = (r+1)/(n)$\frac{^{n-1}C_r}{^nC_{r+1}} = \frac{r+1}{n}$ and use the strict combinatorial limit 0 < (r+1)/(n) ≤ 1$0 < \frac{r+1}{n} \le 1$ to form algebraic inequalities.

Chapter Mix

Class 11 Maths: Binomial Theorem Class 11 Maths: Linear Inequalities

Related Formula

Core Logic

To find the sum of all coefficients in a polynomial expansion, substitute the variable x = 1$x = 1$.

For A$A$ (Sum of coefficients of (1-3x+10x²)ⁿ$(1-3x+10x^2)^n$):

For B$B$ (Sum of coefficients of (1+x²)ⁿ$(1+x^2)^n$):

Step 1: Establishing the Relation

We have A = 8ⁿ$A = 8^n$ and B = 2ⁿ$B = 2^n$. Observe that 8 = 2³$8 = 2^3$.

Substituting B = 2ⁿ$B = 2^n$ into the expression: A = B³$A = B^3$

Pattern Recognition

Summing coefficients of any multi-nomial expansion is trivially simple: just plug in x=1$x=1$ (or x=y=z=1$x=y=z=1$ for multivariable expressions). It collapses the variables, leaving only the arithmetic sum of the coefficients.

Chapter Mix

Class 11 Maths: Binomial Theorem