JEE Fundamental of Mathematics Questions

We can use the Fermat’s Little Theorem i.e. $$a^{p-1} \equiv 1 \pmod{p}$$

Here, $$a = 7$$ and $$p = 23$$. Since 23 is prime and 7 is not a multiple of 23, we can apply the theorem:

$$7^{22} \equiv 1 \pmod{23}$$

Now, we have to break down 103 as a multiple of 22. 

$$103 = (22 \times 4) + 15$$

$$\therefore\ $$ $$7^{103}=7^{22\times4+15}=\left(7^{22}\right)^{^4}\times7^{15}$$

We know that $$7^{22} \equiv 1 \pmod{23}$$.

$$7^{103} \equiv 1^4 \cdot 7^{15} \equiv 7^{15} \pmod{23}$$

Now, we need to calculate: $$7^{15}$$ mod 23

We know $$7^2=49\ ≡\ 3$$ (mod 23)

Similarly,  $$7^4=49\ =3^2≡9$$(mod 23)

$$7^8=49\ =9^2=81≡12$$(mod 23)

Now, $$7^{15}=7^8\times7^4\times7^2\times7$$

Substituting the values: $$12\times9\times3\times7$$

$$12\times9=108≡16$$ (mod23)

$$16\times3=48≡2$$ (mod23)

$$2\times7=14$$ (mod23)

Hence, the final answer is 14. 

We need to determine whether $$25^{190} - 19^{190} - 8^{190} + 2^{190}$$ is divisible by 14, 34, both, or neither.

Note that $$14 = 2 \times 7$$ and $$34 = 2 \times 17$$. So we need to check divisibility by 2, 7, and 17.

To begin,

$$25^{190}$$ is odd, $$19^{190}$$ is odd, $$8^{190}$$ is even, $$2^{190}$$ is even.

So $$25^{190} - 19^{190} - 8^{190} + 2^{190} = (\text{odd} - \text{odd}) - (\text{even} - \text{even}) = \text{even} - \text{even} = \text{even}$$.

The expression is divisible by 2.

Next,

Regroup: $$(25^{190} - 8^{190}) - (19^{190} - 2^{190})$$.

Since $$25 - 8 = 17$$, by the algebraic identity $$a^n - b^n$$ is divisible by $$(a - b)$$, we get $$17 \mid (25^{190} - 8^{190})$$.

Since $$19 - 2 = 17$$, similarly $$17 \mid (19^{190} - 2^{190})$$.

Therefore $$17 \mid \left[(25^{190} - 8^{190}) - (19^{190} - 2^{190})\right]$$.

From this,

Since the expression is divisible by both 2 and 17, and $$\gcd(2, 17) = 1$$, the expression is divisible by $$2 \times 17 = 34$$.

Continuing,

We check modulo 7:

$$25 \equiv 4 \pmod{7}$$, $$19 \equiv 5 \pmod{7}$$, $$8 \equiv 1 \pmod{7}$$, $$2 \equiv 2 \pmod{7}$$.

By Fermat's little theorem, $$a^6 \equiv 1 \pmod{7}$$ for $$\gcd(a, 7) = 1$$.

Since $$190 = 6 \times 31 + 4$$, we have $$a^{190} \equiv a^4 \pmod{7}$$.

$$4^4 = 256 \equiv 4 \pmod{7}$$

$$5^4 = 625 \equiv 2 \pmod{7}$$

$$1^4 = 1 \pmod{7}$$

$$2^4 = 16 \equiv 2 \pmod{7}$$

So the expression $$\equiv 4 - 2 - 1 + 2 = 3 \pmod{7}$$.

Since $$3 \not\equiv 0 \pmod{7}$$, the expression is not divisible by 7, and hence not divisible by 14.

Conclusion: The expression is divisible by 34 but not by 14.

The correct answer is Option C.

We need to find the remainder when $$(2023)^{2023}$$ is divided by $$35$$.

Since $$35 = 5 \times 7$$ and $$\gcd(5, 7) = 1$$, we use the Chinese Remainder Theorem.

Find $$(2023)^{2023} \pmod{5}$$.

$$2023 = 404 \times 5 + 3$$, so $$2023 \equiv 3 \pmod{5}$$.

By Fermat's Little Theorem, $$3^4 \equiv 1 \pmod{5}$$.

$$2023 = 4 \times 505 + 3$$, so $$2023 \equiv 3 \pmod{4}$$.

Therefore: $$3^{2023} \equiv 3^3 = 27 \equiv 2 \pmod{5}$$.

Find $$(2023)^{2023} \pmod{7}$$.

$$2023 = 289 \times 7$$, so $$2023 \equiv 0 \pmod{7}$$.

Therefore: $$(2023)^{2023} \equiv 0 \pmod{7}$$.

Apply the Chinese Remainder Theorem.

We need $$x$$ such that:

$$x \equiv 2 \pmod{5}$$ and $$x \equiv 0 \pmod{7}$$.

From the second condition, $$x = 7k$$ for some integer $$k$$.

Substituting: $$7k \equiv 2 \pmod{5} \implies 2k \equiv 2 \pmod{5} \implies k \equiv 1 \pmod{5}$$.

So $$k = 5m + 1$$ and $$x = 7(5m + 1) = 35m + 7$$.

The remainder when $$(2023)^{2023}$$ is divided by $$35$$ is $$\boxed{7}$$.

We need $$3^n - 3 \equiv 0 \pmod{7}$$, i.e., $$3^n \equiv 3 \pmod{7}$$, i.e., $$3^{n-1} \equiv 1 \pmod{7}$$.

The order of 3 modulo 7: $$3^1 = 3$$, $$3^2 = 9 \equiv 2$$, $$3^3 = 27 \equiv 6$$, $$3^4 = 81 \equiv 4$$, $$3^5 = 243 \equiv 5$$, $$3^6 = 729 \equiv 1 \pmod{7}$$.

So $$3^{n-1} \equiv 1 \pmod{7}$$ when $$6 | (n-1)$$, i.e., $$n \equiv 1 \pmod{6}$$.

Values of $$n$$ in $$[10, 100]$$ with $$n \equiv 1 \pmod{6}$$:

$$n = 13$$, $$19$$, $$25$$, $$31$$, $$37$$, $$43$$, $$49$$, $$55$$, $$61$$, $$67$$, $$73$$, $$79$$, $$85$$, $$91$$, $$97$$

Count: $$\frac{97 - 13}{6} + 1 = \frac{84}{6} + 1 = 14 + 1 = 15$$

The answer is $$\mathbf{15}$$.

We need to find $$5^{99} \mod 11$$.

By Fermat's Little Theorem, since 11 is prime and $$\gcd(5, 11) = 1$$:

$$ 5^{10} \equiv 1 \pmod{11} $$

Dividing the exponent: $$99 = 10 \times 9 + 9$$

$$ 5^{99} = (5^{10})^9 \times 5^9 \equiv 1^9 \times 5^9 = 5^9 \pmod{11} $$

Computing $$5^9 \mod 11$$ step by step:

$$5^1 \equiv 5$$

$$5^2 \equiv 25 \equiv 3 \pmod{11}$$

$$5^4 \equiv 3^2 = 9 \pmod{11}$$

$$5^8 \equiv 9^2 = 81 \equiv 4 \pmod{11}$$

$$5^9 = 5^8 \times 5 \equiv 4 \times 5 = 20 \equiv 9 \pmod{11}$$

The remainder is 9.

We need to find the remainder when $$(2021)^{2023}$$ is divided by 7.

Find $$2021 \mod 7$$.

$$2021 = 288 \times 7 + 5$$

So $$2021 \equiv 5 \pmod{7}$$

Find the pattern of powers of 5 modulo 7.

$$5^1 \equiv 5 \pmod{7}$$

$$5^2 \equiv 25 \equiv 4 \pmod{7}$$

$$5^3 \equiv 20 \equiv 6 \pmod{7}$$

$$5^4 \equiv 30 \equiv 2 \pmod{7}$$

$$5^5 \equiv 10 \equiv 3 \pmod{7}$$

$$5^6 \equiv 15 \equiv 1 \pmod{7}$$

The powers of 5 repeat with a cycle of length 6 (by Fermat's Little Theorem).

Find $$2023 \mod 6$$.

$$2023 = 337 \times 6 + 1$$

So $$2023 \equiv 1 \pmod{6}$$

Compute the remainder.

$$(2021)^{2023} \equiv 5^{2023} \equiv 5^1 \equiv 5 \pmod{7}$$

The remainder is $$\textbf{5}$$.

The correct answer is Option D.

We want to determine the remainder when $$(11)^{1011} + (1011)^{11}$$ is divided by 9. First, observe that $$11 \equiv 2 \pmod{9}$$ so that $$11^{1011} \equiv 2^{1011} \pmod{9}$$. To compute this, note that the powers of 2 modulo 9 cycle with period 6:

$$2^1 \equiv 2, \quad 2^2 \equiv 4, \quad 2^3 \equiv 8, \quad 2^4 \equiv 7, \quad 2^5 \equiv 5, \quad 2^6 \equiv 1 \pmod{9}$$

Since $$1011 = 6 \times 168 + 3$$, it follows that $$2^{1011} \equiv 2^3 \equiv 8 \pmod{9}$$.

Next, we reduce $$1011^{11}$$ modulo 9. Writing $$1011 = 112 \times 9 + 3$$ shows that $$1011 \equiv 3 \pmod{9}$$, hence $$1011^{11} \equiv 3^{11} \pmod{9}$$. Because $$3^2 = 9 \equiv 0 \pmod{9}$$, one factor of $$3^2$$ makes the product a multiple of 9, so $$3^{11} = 3^2 \times 3^9 = 9 \times 3^9 \equiv 0 \pmod{9}$$.

Adding these results gives

$$11^{1011} + 1011^{11} \equiv 8 + 0 \equiv 8 \pmod{9}$$

The correct answer is Option B: $$8$$.

$$3^{2022} = (3^2)^{1011}$$

$$(3^2)^{1011} \equiv (-1)^{1011} \pmod{5}$$

$$(-1)^{1011} = -1$$

Now we know that a remainder of $$-1$$ is not the standard way. So to convert a negative remainder back to a positive one, we can just add the divisor ($$5$$):

Hence, $$3^{2022} \equiv 4 \pmod{5}$$

Alternative Method 1:

Using Fermat's Little Theorem: $$a^{p-1} \equiv 1 \pmod{p}$$

So, we can say, $$3^{5-1} \equiv 3^4 \equiv 1 \pmod{5}$$

Now, $$2022 = 4 \times 505 + 2$$

Hence, we can say, $$3^{2022} = 3^{4 \times 505 + 2} = (3^4)^{505} \times 3^2$$

$$3^{2022} \equiv (1)^{505} \times 3^2 \pmod{5}$$

$$3^{2022} \equiv 1 \times 9 \pmod{5}$$

$$3^{2022} \equiv 4 \pmod{5}$$

Alternative Method 2:

We can also find the pattern. 

$$3^1 \equiv 3 \pmod{5}$$

$$3^2 = 9 \equiv 4 \pmod{5}$$

$$3^3 = 27 \equiv 2 \pmod{5}$$

$$3^4 = 81 \equiv 1 \pmod{5}$$

$$3^5 = 243 \equiv 3 \pmod{5}$$

So the cycle repeats: 3, 4, 2, 1

Exponent is 2022. 

Hence, $$2022 \div 4 = 505 \text{ with a remainder of } 2$$

Since the remainder is $$2$$, the result would be same as the second position in our cycle ($$3^2 \equiv 4 \pmod{5}$$).

We are given $$n = 2^x 3^y 5^z$$ where $$y + z = 5$$ and $$y^{-1} + z^{-1} = \dfrac{5}{6}$$, with $$y > z$$.

From the second equation, $$\dfrac{1}{y} + \dfrac{1}{z} = \dfrac{y + z}{yz} = \dfrac{5}{yz} = \dfrac{5}{6}$$, which gives $$yz = 6$$.

So $$y$$ and $$z$$ are roots of $$t^2 - 5t + 6 = 0$$, i.e., $$(t - 3)(t - 2) = 0$$. Since $$y > z$$, we get $$y = 3$$ and $$z = 2$$.

Thus $$n = 2^x \cdot 3^3 \cdot 5^2$$. The odd divisors of $$n$$ are the divisors of $$3^3 \cdot 5^2$$ (we exclude all factors of 2). The number of such divisors is $$(3 + 1)(2 + 1) = 12$$.

The answer is $$12$$.

We need the remainder when $$(2021)^{3762}$$ is divided by 17.

First, $$2021 \mod 17$$: since $$17 \times 118 = 2006$$, we get $$2021 - 2006 = 15$$, so $$2021 \equiv 15 \equiv -2 \pmod{17}$$.

Since the exponent 3762 is even, $$(2021)^{3762} \equiv (-2)^{3762} = 2^{3762} \pmod{17}$$.

By Fermat's Little Theorem (17 is prime, $$\gcd(2,17)=1$$): $$2^{16} \equiv 1 \pmod{17}$$.

Dividing: $$3762 = 16 \times 235 + 2$$, so $$2^{3762} = (2^{16})^{235} \cdot 2^2 \equiv 1 \cdot 4 = 4 \pmod{17}$$.

The remainder is 4.

We are asked to consider all natural numbers $$n$$ that satisfy the double inequality $$100<n<200$$ and whose highest common factor with $$91$$ is greater than $$1$$.

First, we factorise $$91$$ so that we know which prime numbers can possibly divide both $$91$$ and $$n$$.

We have $$91 = 7 \times 13$$, and $$7$$ and $$13$$ are primes.

If the H.C.F. (also called G.C.D.) of $$91$$ and $$n$$ is to be greater than $$1$$, then $$n$$ must share at least one of the prime factors $$7$$ or $$13$$ with $$91$$. Equivalently, $$n$$ must be divisible by $$7$$ or by $$13$$ (or by both).

So we now list every multiple of $$7$$ and every multiple of $$13$$ that lies strictly between $$100$$ and $$200$$.

Multiples of $$7$$ in the required interval
We solve $$7k$$ with $$100<7k<200$$.

Dividing by $$7$$ gives $$\dfrac{100}{7}<k<\dfrac{200}{7}$$, i.e. $$14.28\dots<k<28.57\dots$$. Hence the integer values of $$k$$ are $$k=15,16,\dots ,28$$. Writing the corresponding numbers:

$$\begin{aligned} 7\times15 &= 105 \\ 7\times16 &= 112 \\ 7\times17 &= 119 \\ 7\times18 &= 126 \\ 7\times19 &= 133 \\ 7\times20 &= 140 \\ 7\times21 &= 147 \\ 7\times22 &= 154 \\ 7\times23 &= 161 \\ 7\times24 &= 168 \\ 7\times25 &= 175 \\ 7\times26 &= 182 \\ 7\times27 &= 189 \\ 7\times28 &= 196 \end{aligned}$$

So the complete list of admissible multiples of $$7$$ is
$$105,112,119,126,133,140,147,154,161,168,175,182,189,196.$$

Multiples of $$13$$ in the required interval
We solve $$13m$$ with $$100<13m<200$$.

Dividing by $$13$$ gives $$\dfrac{100}{13}<m<\dfrac{200}{13}$$, i.e. $$7.69\dots<m<15.38\dots$$. Hence the integer values of $$m$$ are $$m=8,9,\dots ,15$$. Writing the corresponding numbers:

$$\begin{aligned} 13\times8 &= 104 \\ 13\times9 &= 117 \\ 13\times10 &= 130 \\ 13\times11 &= 143 \\ 13\times12 &= 156 \\ 13\times13 &= 169 \\ 13\times14 &= 182 \\ 13\times15 &= 195 \end{aligned}$$

So the complete list of admissible multiples of $$13$$ is
$$104,117,130,143,156,169,182,195.$$

Combining the two lists
Every number in either list satisfies the H.C.F. condition, but we must beware of double counting any number that is simultaneously a multiple of $$7$$ and $$13$$.

Notice that $$182$$ appears in both lists because $$182 = 7\times26 = 13\times14$$ (since $$182$$ is itself $$91\times2$$). No other number is common to the two lists.

Sum of all admissible numbers

First sum all the multiples of $$7$$ we found:

$$\begin{aligned} &105+112+119+126+133+140+147+154+161+168+175+182+189+196 \\ &= (105+112) + (119+126) + (133+140) + (147+154) + (161+168) + (175+182) + (189+196) \\ &= 217 + 245 + 273 + 301 + 329 + 357 + 385 \\ &= 2107 \end{aligned}$$

Next sum all the multiples of $$13$$ we found:

$$\begin{aligned} &104+117+130+143+156+169+182+195 \\ &= (104+117) + (130+143) + (156+169) + (182+195) \\ &= 221 + 273 + 325 + 377 \\ &= 1196 \end{aligned}$$

Because $$182$$ lies in both lists, it has been counted twice, once in each subtotal. To obtain the correct total we subtract this overlap once.

Therefore the required sum is

$$2107 + 1196 - 182 = 3121.$$

Hence, the correct answer is Option C.

We want the fractional part of $$\dfrac{2^{403}}{15}$$. By definition, if we divide $$2^{403}$$ by $$15$$ we can write

$$2^{403}=15q+k$$

where $$q$$ is the quotient and $$k$$ is the remainder satisfying $$0\le k<15$$. Then

$$\dfrac{2^{403}}{15}=q+\dfrac{k}{15},$$

so the fractional part is exactly $$\dfrac{k}{15}$$. Our task is therefore to find the remainder $$k$$ when $$2^{403}$$ is divided by $$15$$; symbolically, we must evaluate

$$k\equiv 2^{403}\pmod{15}.$$

To do this efficiently we recall Euler’s theorem. First we note that $$\gcd(2,15)=1$$, so Euler’s theorem applies. The theorem states:

$$\text{If }\gcd(a,n)=1,\text{ then }a^{\phi(n)}\equiv1\pmod n.$$

Here $$n=15$$, so we compute Euler’s totient function $$\phi(15)$$. Since $$15=3\cdot5$$ with both factors prime, we have

$$\phi(15)=15\left(1-\dfrac13\right)\left(1-\dfrac15\right)=15\cdot\dfrac23\cdot\dfrac45=8.$$

Therefore, by Euler’s theorem,

$$2^{8}\equiv1\pmod{15}.$$

Now we use this congruence to break the large exponent $$403$$ into manageable parts. Write the exponent in the form

$$403=8\cdot50+3.$$

So,

$$2^{403}=2^{8\cdot50+3}=2^{8\cdot50}\cdot2^{3}=\left(2^{8}\right)^{50}\cdot2^{3}.$$

Since $$2^{8}\equiv1\pmod{15},$$ raising it to any power keeps it congruent to $$1$$:

$$\left(2^{8}\right)^{50}\equiv1^{50}\equiv1\pmod{15}.$$

Hence,

$$2^{403}\equiv1\cdot2^{3}\pmod{15}.$$

We still need $$2^{3}$$, and that is straightforward:

$$2^{3}=8.$$

So we finally have

$$2^{403}\equiv8\pmod{15}.$$

Comparing this with our earlier definition $$2^{403}=15q+k,$$ we see the remainder is

$$k=8.$$

Because $$0\le k<15,$$ this value is valid. The fractional part of $$\dfrac{2^{403}}{15}$$ is therefore $$\dfrac{8}{15},$$ and the asked integer is $$k=8$$.

Hence, the correct answer is Option C.

We are given three statements for integers $$m$$ and $$n$$, both greater than 1:

• P: $$m$$ divides $$n$$ (i.e., $$n$$ is divisible by $$m$$, written as $$m \mid n$$)
• Q: $$m$$ divides $$n^2$$ (i.e., $$m \mid n^2$$)
• R: $$m$$ is prime

We need to evaluate the logical implications given in the options and determine which one is always true.

Starting with Option A: $$Q \wedge R \rightarrow P$$. This means if Q and R are both true, then P must be true. That is, if $$m$$ divides $$n^2$$ and $$m$$ is prime, then $$m$$ divides $$n$$.

Assume Q and R are true: $$m \mid n^2$$ and $$m$$ is prime. Since $$m$$ is prime and divides $$n^2$$, and $$n^2 = n \times n$$, by Euclid's lemma (which states that if a prime divides a product, it must divide at least one factor), $$m$$ must divide $$n$$. Therefore, P is true. This implication holds.

Now, Option B: $$P \wedge Q \rightarrow R$$. This means if P and Q are true, then R must be true. That is, if $$m$$ divides $$n$$ and $$m$$ divides $$n^2$$, then $$m$$ is prime.

Note that if P is true ($$m \mid n$$), then Q ($$m \mid n^2$$) is automatically true because if $$n = m \cdot k$$ for some integer $$k$$, then $$n^2 = m \cdot (m \cdot k^2)$$, so $$m \mid n^2$$. Thus, the premise $$P \wedge Q$$ reduces to P being true. But the conclusion requires $$m$$ to be prime, which may not hold. For example, let $$m = 4$$ (composite) and $$n = 4$$. Then P: $$4 \mid 4$$ is true (since $$4 = 4 \times 1$$), and Q: $$4 \mid 16$$ is true (since $$16 = 4 \times 4$$), but R: $$m = 4$$ is not prime. Since the premise is true and the conclusion is false, this implication is not always true.

Option C: $$Q \rightarrow R$$. This means if Q is true, then R must be true. That is, if $$m$$ divides $$n^2$$, then $$m$$ is prime.

This is not always true. For example, let $$m = 4$$ and $$n = 2$$. Then Q: $$4 \mid 2^2 = 4$$ is true (since $$4 = 4 \times 1$$), but R: $$m = 4$$ is not prime. Thus, this implication is false.

Option D: $$Q \rightarrow P$$. This means if Q is true, then P must be true. That is, if $$m$$ divides $$n^2$$, then $$m$$ divides $$n$$.

This is not always true. For example, let $$m = 4$$ and $$n = 2$$. Then Q: $$4 \mid 2^2 = 4$$ is true, but P: $$4 \mid 2$$ is false (since $$2 \div 4 = 0.5$$, not an integer). Another example: $$m = 9$$ and $$n = 3$$. Then Q: $$9 \mid 3^2 = 9$$ is true (since $$9 = 9 \times 1$$), but P: $$9 \mid 3$$ is false. Hence, this implication is false.

Only Option A is always true. Therefore, the correct answer is Option A.

Hence, the correct answer is Option A.

Firstly, we need to find the remainder when 428 is divided by 21.

$$428 = (21 \times 20) + 8$$

So, $$428 \equiv 8 \pmod{21}$$.

Hence, we can say: $$428^{2024} \equiv 8^{2024} \pmod{21}$$

Now, we can find the remainders for the first few powers of 8 so as to find any recurring pattern of the remainders. 

$$8^1\ ≡\ 8$$ (mod 21)

$$8^2=24$$ (mod 21)

Now, find the remainder of 64 divided by 21: $$64 = (21 \times 3) + 1$$

So, $$8^2 \equiv 1 \pmod{21}$$.

Now, because $$8^2 \equiv 1 \pmod{21}$$, we can express the large exponent, 2024, in terms of 2. We divide 2024 by 2: 

$$2024 = 2 \times 1012$$

So, we can say, $$8^{2024}=\left(8^2\right)^{2024}$$

Substitute 1 for $$8^2$$,

$$(8^2)^{1012} \equiv 1^{1012} \pmod{21}$$

Hence, the remainder wil be 1. 

We are given that the $$50^{th}$$ root of $$x$$ is $$12$$ and the $$50^{th}$$ root of $$y$$ is $$18$$.

So $$x = 12^{50}$$ and $$y = 18^{50}$$.

By Euler's theorem, since $$\gcd(12, 25) = 1$$ and $$\phi(25) = 20$$, we have $$12^{20} \equiv 1 \pmod{25}$$.

$$12^{50} = 12^{20 \cdot 2} \cdot 12^{10} = (12^{20})^2 \cdot 12^{10} \equiv 12^{10} \pmod{25}$$

Now compute $$12^{10} \pmod{25}$$:

$$12^2 = 144 \equiv 19 \pmod{25}$$

$$12^4 = 19^2 = 361 \equiv 11 \pmod{25}$$

$$12^8 = 11^2 = 121 \equiv 21 \pmod{25}$$

$$12^{10} = 12^8 \cdot 12^2 = 21 \times 19 = 399 \equiv 24 \pmod{25}$$

$$18^2 = 324 \equiv 24 \equiv -1 \pmod{25}$$

$$18^{50} = (18^2)^{25} = (-1)^{25} = -1 \equiv 24 \pmod{25}$$

$$(x + y) \pmod{25} = (24 + 24) \pmod{25} = 48 \pmod{25} = 23$$

The remainder is $$\boxed{23}$$.

$$22^{2022} + 2022^{22} \pmod{3}$$:

$$22 \equiv 1 \pmod{3}$$, so $$22^{2022} \equiv 1$$. $$2022 \equiv 0 \pmod{3}$$, so $$2022^{22} \equiv 0$$.

$$\alpha = 1 + 0 = 1$$.

$$22^{2022} + 2022^{22} \pmod{7}$$:

$$22 \equiv 1 \pmod{7}$$, so $$22^{2022} \equiv 1$$. $$2022 = 288 \times 7 + 6$$, so $$2022 \equiv -1 \pmod{7}$$, $$2022^{22} \equiv 1$$.

$$\beta = 1 + 1 = 2$$.

$$\alpha^2 + \beta^2 = 1 + 4 = 5$$.

The correct answer is Option 3: 5.

To obtain the product of the last two digits of $$(1919)^{1919}$$ we must first find those two digits, i.e. evaluate $$(1919)^{1919} \pmod{100}$$.

Step 1 : Reduce the base modulo 100
$$1919 \equiv 19 \pmod{100}$$
Hence $$ (1919)^{1919} \equiv 19^{1919} \pmod{100} $$.

Step 2 : Reduce the exponent using Euler’s theorem
Since $$\gcd(19,100)=1$$, Euler’s theorem gives $$19^{\phi(100)} \equiv 1 \pmod{100}$$.
Factorising $$100 = 2^{2}\cdot5^{2}$$, $$\phi(100)=100\left(1-\frac{1}{2}\right)\left(1-\frac{1}{5}\right)=40$$.

Therefore the powers of 19 repeat every 40:
$$19^{40} \equiv 1 \pmod{100}$$.

Now reduce the exponent 1919 modulo 40:
$$1919 = 40\times47 + 39 \quad\Longrightarrow\quad 1919 \equiv 39 \pmod{40}$$.

Thus
$$19^{1919} \equiv 19^{39} \pmod{100} \quad -(1)$$

Step 3 : Shorten the exponent further (optional but quicker)
We shall show that $$19^{10} \equiv 1 \pmod{100}$$, which will cut the power down even more.

Compute successive powers of 19 (always keeping only the last two digits):
$$19^{1}=19$$
$$19^{2}=361\equiv61$$
$$19^{3}=61\cdot19=1159\equiv59$$
$$19^{4}=59\cdot19=1121\equiv21$$
$$19^{5}=21\cdot19=399\equiv99$$
$$19^{6}=99\cdot19=1881\equiv81$$
$$19^{7}=81\cdot19=1539\equiv39$$
$$19^{8}=39\cdot19=741\equiv41$$
$$19^{9}=41\cdot19=779\equiv79$$
$$19^{10}=79\cdot19=1501\equiv01\equiv1$$

Since $$19^{10}\equiv1\pmod{100}$$, rewrite the required power:

$$19^{39}=19^{10\cdot3}\cdot19^{9}\equiv(19^{10})^{3}\cdot19^{9}\equiv1^{3}\cdot19^{9}\equiv19^{9}\pmod{100}$$.

Step 4 : Evaluate $$19^{9}\pmod{100}$$ (already calculated above)
From the list, $$19^{9}\equiv79\pmod{100}$$.
Hence the last two digits of $$(1919)^{1919}$$ are 7 and 9.

Step 5 : Required product
Product of the digits $$7 \times 9 = 63$$.

Final Answer : 63

To find: the remainder of  $$64^{32^{32}}$$

We know $$9\times\ 7=63$$

$$\therefore$$ When we divide $$\frac{64}{7}$$ we get remainder 1.

So we can also write $$64^{32^{32}}$$ as $$1^{32^{32}}$$

When we divide $$1^{32^{32}}$$ by 9 we get a remainder 1.

$$\therefore$$ correct answer is 1.

ALTERNATE METHOD:

Simplify 64 mod 9

$$64=\left(9\times\ 7\right)+1$$

So, 64 ≡1 (mod 9) 

Now raise both side to the power $$32^{32}$$

$$\therefore$$ $$64^{32^{32}}≡1^{32^{32}}$$ (mod 9)

Since any power of 1 is 1 

$$\therefore$$ $$64^{32^{32}}$$ is divided by 9 the remainder is 1.