JEE Number System 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. 

Let the digits be $$a, b, c$$ , where  $$a \ge 2$$, and $$b, c \ge 0$$, and all ≤ 9.

We want $$ a + b + c = 15 $$ with the number between 212 and 999.

Let, $$ a = 2$$

If $$a = 2$$ , then the number must be ≥ 212, so $$b \ge 1$$

And, $$ b + c = 13$$

Valid pairs are $$ (4,9),(5,8),(6,7),(7,6),(8,5),(9,4) $$

That’s 6 pairs, all satisfy $$ b \ge 1$$ So, 6 numbers are obtained in this case. 

Similarly, we fix for other values of a.

a = 3, b+c = 12

Possible choices: (3,9) to (9,3) → 7 numbers

a = 4 , b+c = 11

Possible choices: (2,9) to (9,2) → 8 numbers

a = 5 , b+c = 10

Possible choices: (1,9) to (9,1) → 9 numbers

a = 6 , b+c = 9

Possible choices: (0,9) to (9,0) → 10 numbers

a = 7 , b+c = 8

Possible choices: (0,8) to (8,0) → 9 numbers

a = 8 , b+c = 7

Possible choices: (0,7) to (7,0) → 8 numbers

a = 9 , b+c = 6

Possible choices: (0,6) to (6,0) → 7 numbers

So, the total possible choices of numbers: 6 + 7 + 8 + 9 + 10 + 9 + 8 + 7 = 64

Every 2-digit number having greatest common divisor $$6$$ with another 2-digit number must itself be divisible by $$6$$, because any common divisor of the two numbers divides their gcd.
Hence write the numbers as $$m = 6a$$ and $$n = 6b$$ with $$a,b \in \mathbb{N}$$.

Because $$m$$ and $$n$$ are 2-digit numbers,

$$10 \le m = 6a \le 99 \quad \text{and} \quad 10 \le n = 6b \le 99$$

which gives

$$\frac{10}{6} \le a \le \frac{99}{6}, \qquad \frac{10}{6} \le b \le \frac{99}{6}$$

or, after taking integer values,

$$2 \le a \le 16, \qquad 2 \le b \le 16.$$

The condition $$\gcd(m,n)=6$$ becomes $$\gcd(6a,6b)=6,$$ i.e.

$$\gcd(a,b)=1.$$ Thus we must count ordered pairs $$(a,b)$$ such that

$$2 \le a \lt b \le 16 \quad \text{and} \quad \gcd(a,b)=1.$$ Call such pairs “coprime pairs”.

List each value of $$a$$ and count suitable $$b$$ values individually:

$$\begin{aligned} a=2 &: b=3,5,7,9,11,13,15 &\Rightarrow 7\\ a=3 &: b=4,5,7,8,10,11,13,14,16 &\Rightarrow 9\\ a=4 &: b=5,7,9,11,13,15 &\Rightarrow 6\\ a=5 &: b=6,7,8,9,11,12,13,14,16 &\Rightarrow 9\\ a=6 &: b=7,11,13 &\Rightarrow 3\\ a=7 &: b=8,9,10,11,12,13,15,16 &\Rightarrow 8\\ a=8 &: b=9,11,13,15 &\Rightarrow 4\\ a=9 &: b=10,11,13,14,16 &\Rightarrow 5\\ a=10&: b=11,13 &\Rightarrow 2\\ a=11&: b=12,13,14,15,16 &\Rightarrow 5\\ a=12&: b=13 &\Rightarrow 1\\ a=13&: b=14,15,16 &\Rightarrow 3\\ a=14&: b=15 &\Rightarrow 1\\ a=15&: b=16 &\Rightarrow 1 \end{aligned}$$

Adding these counts:

$$7+9+6+9+3+8+4+5+2+5+1+3+1+1 = 64.$$

Therefore there are $$64$$ ordered pairs $$(a,b)$$, and hence $$64$$ ordered pairs $$(m,n)$$ with $$m \lt n$$ and $$\gcd(m,n)=6$$ where both $$m$$ and $$n$$ are 2-digit numbers.

Final answer: $$64$$.

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

We have, $$4!=24$$, $$5!=120$$, and $$3!=6$$

Therefore,

$$\alpha = \dfrac{24!}{24^6}$$  and  $$\beta = \dfrac{120!}{120^{24}}$$

Or

$$\alpha = \dfrac{24!}{2^{18}\times 3^6}$$  and  $$\beta = \dfrac{120!}{2^{72}\times 3^{24}\times 5^{24}}$$

In order for $$\alpha$$ to be an integer, $$24!$$ must contain $$2^{18}$$ and $$3^6$$

The highest power of $$2$$ in $$24!$$ is $$\lfloor \dfrac{24}{2} \rfloor + \lfloor \dfrac{24}{4} \rfloor + \lfloor \dfrac{24}{8} \rfloor + \lfloor \dfrac{24}{16} \rfloor + \dots = 12+6+3+1= 22$$ which is larger than $$18$$

Similarly, the highest power of $$3$$ in $$24!$$ is larger than $$6$$.

Thus, $$\alpha \in \mathbb{N}$$

In $$\beta$$, in order for $$\beta$$ to be an integer, $$120!$$ must contain $$2^{72}$$, $$3^{24}$$, and $$5^{24}$$

The highest power of $$2$$ in $$120!$$ is $$60+30+15+7+3+1= 116$$, which is larger than $$72$$. The highest power of $$3$$ in $$120!$$ is $$40+13+4+1= 58$$, which is also larger than $$24$$. The highest power of $$5$$ in $$120!$$ is $$24+4= 28$$, which is larger than $$2$$.

Thus, $$\beta \in \mathbb{N}$$

Option C is the correct answer.

Total 3-digit numbers (no repetition): $$^5P_3=5\times4\times3=60$$

Now, for a number to be divisible by 3, the sum of its digits must be divisible by 3.

So, classifying digits by mod 3:

Mod 0: {3}

Mod 1: {4, 7}

Mod 2: {2, 5}

For sum ≡ 0 (mod 3), possible valid combination:

(0, 1, 2)

Number of ways of choosing 1 from each group:   $$1\times2\times2=4$$ ways (digit sets)

Each set can be arranged in: $$3!=6$$ ways

So total numbers divisible by 3:$$4\times6=24$$

We need to count 3-digit integers (100-999) whose digit sum equals 14.

Let the three-digit number have digits $$a$$ (hundreds), $$b$$ (tens), $$c$$ (units) where $$1 \leq a \leq 9$$, $$0 \leq b \leq 9$$, $$0 \leq c \leq 9$$, and $$a + b + c = 14$$.

Since $$a \geq 1$$, set $$a' = a - 1$$ so that $$0 \leq a' \leq 8$$ and the equation becomes $$a' + b + c = 13$$.

Without considering the upper bounds on $$a'$$, $$b$$, and $$c$$, the number of nonnegative integer solutions is given by stars and bars as $$\binom{13 + 2}{2} = \binom{15}{2} = 105$$.

To account for the constraints, subtract cases where one of the variables exceeds its bound using inclusion-exclusion.

If $$a' \geq 9$$, let $$a'' = a' - 9$$, giving $$a'' + b + c = 4$$ and $$\binom{4 + 2}{2} = 15$$ solutions. If $$b \geq 10$$, let $$b' = b - 10$$, giving $$a' + b' + c = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. If $$c \geq 10$$, let $$c' = c - 10$$, giving $$a' + b + c' = 3$$ and $$\binom{3 + 2}{2} = 10$$ solutions. Any intersection of two or more such violations is impossible because it would require a sum exceeding 13.

Hence the total count is $$\text{Count} = 105 - 15 - 10 - 10 = 70$$.

The answer is 70.

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. 

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.

Let $$N_r$$ denote the number whose first and last digits are 7 and whose $$r$$ middle digits are all 5. Thus the total number of digits is $$(r+2)$$ and

$$N_r = 7\cdot 10^{\,r+1} + 5\Bigl(10^{\,r-1}+10^{\,r-2}+\dots+10^1\Bigr) + 7.$$

The parenthesis is a geometric series of $$r$$ terms with first term $$10^{\,r-1}$$ and ratio 1/10, whose sum equals $$\dfrac{10^{\,r}-1}{9} \times 10.$$ Therefore

$$N_r \;=\; 7\cdot 10^{r+1} \;+\; \dfrac{50}{9}\,(10^{\,r}-1) \;+\; 7.$$

The required sum is

$$S \;=\; \sum_{r=0}^{98} N_r.$$ We add $$S$$ term-wise by separating coefficient-wise parts.

First part (leading 7): $$S_1 = 7\sum_{r=0}^{98}10^{\,r+1}=70\sum_{r=0}^{98}10^{\,r} = 70\cdot\dfrac{10^{99}-1}{9}.$$

Second part (middle 5’s): $$S_2 = \dfrac{50}{9}\sum_{r=0}^{98}(10^{\,r}-1) = \dfrac{50}{9}\Bigl(\dfrac{10^{99}-1}{9}-99\Bigr) = \dfrac{50}{81}\bigl(10^{99}-892\bigr).$$

Third part (trailing 7): $$S_3 = 7\times 99 = 693.$$

Putting $$A = 10^{99},$$ we get

$$S = S_1+S_2+S_3 = \dfrac{70A}{9}-\dfrac{70}{9}+\dfrac{50A}{81}-\dfrac{50\!\cdot\!892}{81}+693 = \dfrac{680A + 10\,903}{81}.$$

Next write the term with 99 middle 5’s, i.e. $$N_{99}$$ (the number $$\overset{99}{75\ldots57}$$).

Since $$10^{100}=10A,$$ we have

$$N_{99}=7\cdot10^{100}+\dfrac{50}{9}(10^{99}-1)+7 = 70A+\dfrac{50A-50}{9}+7 = \dfrac{680A+13}{9}.$$

We are told that

$$S = \dfrac{\,N_{99}+m\,}{n}, \qquad m,n\in\mathbb{N},\;m,n\lt3000.$$ Substituting the obtained expressions:

$$\dfrac{680A+10\,903}{81} = \dfrac{\dfrac{680A+13}{9}+m}{n}.$$

Cross-multiplying,

$$n(680A + 10\,903) = 9(680A + 13) + 81m.$$

Equating coefficients of $$A$$ gives $$680n = 9\!\times\!680 \;\Longrightarrow\; n = 9.$$

Equating the constant terms then yields

$$10\,903n = 117 + 81m \;\Longrightarrow\; 10\,903\!\times\!9 = 117 + 81m \;\Longrightarrow\; 98\,127 = 117 + 81m \;\Longrightarrow\; 81m = 98\,010 \;\Longrightarrow\; m = 1\,210.$$

Both $$m$$ and $$n$$ are below 3000, so they satisfy the given restriction.

Finally, $$m + n = 1\,210 + 9 = 1\,219.$$

Hence the required value is 1219.

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

$$22 \equiv 1 \pmod{3}$$, 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.

We need to check the validity of two statements.

First, we verify that $$2023^{2022} - 1999^{2022}$$ is divisible by 8. Observing that $$2023 \equiv 7 \equiv -1 \pmod{8}$$ and $$1999 \equiv 7 \equiv -1 \pmod{8},$$ and noting that 2022 is even, we obtain the congruences

$$2023^{2022} \equiv (-1)^{2022} = 1 \pmod{8}$$

$$1999^{2022} \equiv (-1)^{2022} = 1 \pmod{8}$$

It follows that $$2023^{2022} - 1999^{2022} \equiv 0 \pmod{8}$$, so the first statement is true.

Next, we consider whether $$13 \cdot 13^n - 11n - 13$$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$. Defining $$f(n) = 13^{n+1} - 11n - 13$$, we use the binomial expansion $$13^{n+1} = 13(1+12)^n$$ to write

$$13^{n+1} = 13\left[1 + 12n + \binom{n}{2}144 + \binom{n}{3}12^3 + \cdots\right]$$

$$= 13 + 156n + 13\binom{n}{2} \cdot 144 + \text{(higher terms divisible by 144)}$$

Subtracting $$11n + 13$$ yields

$$f(n) = 13 + 156n + 13\binom{n}{2} \cdot 144 + \cdots - 11n - 13 = 145n + 144 \cdot (\text{integer})$$

Hence $$f(n) \equiv 145n \equiv n \pmod{144}$$, and $$f(n)$$ is divisible by 144 precisely when $$n$$ is divisible by 144. Since there are infinitely many multiples of 144 in $$\mathbb{N}$$, the second statement is also true.

Both statements are correct, and the answer is Option C: Both (S1) and (S2) are correct.

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 $$\boxed{7}$$.

Find the total number of 4-digit numbers whose GCD with 54 is 2.

$$54 = 2 \times 3^3$$

$$\gcd(n, 54) = 2$$ means:

- $$n$$ is divisible by 2 (even), and

- $$\gcd(n, 27) = 1$$, i.e., $$n$$ is not divisible by 3.

4-digit numbers range from 1000 to 9999. Total = 9000.

Even 4-digit numbers: $$\frac{9000}{2} = 4500$$

Multiples of 6 from 1000 to 9999: $$\lfloor 9999/6 \rfloor - \lfloor 999/6 \rfloor = 1666 - 166 = 1500$$

Count = $$4500 - 1500 = 3000$$

The answer is $$\boxed{3000}$$.

We need to count 4-digit numbers $$\leq 2800$$ divisible by 3 or by 11.

4-digit numbers range from 1000 to 2800.

Using inclusion-exclusion: $$|A \cup B| = |A| + |B| - |A \cap B|$$

Divisible by 3 (set A):

From 1000 to 2800: First multiple of 3 $$\geq 1000$$ is 1002. Last $$\leq 2800$$ is 2799.

Count = $$\frac{2799 - 1002}{3} + 1 = \frac{1797}{3} + 1 = 599 + 1 = 600$$

Divisible by 11 (set B):

First multiple of 11 $$\geq 1000$$: $$\lceil 1000/11 \rceil = 91$$, so $$91 \times 11 = 1001$$.

Last $$\leq 2800$$: $$\lfloor 2800/11 \rfloor = 254$$, so $$254 \times 11 = 2794$$.

Count = $$254 - 91 + 1 = 164$$

Divisible by 33 (set A ∩ B):

First $$\geq 1000$$: $$\lceil 1000/33 \rceil = 31$$, so $$31 \times 33 = 1023$$.

Last $$\leq 2800$$: $$\lfloor 2800/33 \rfloor = 84$$, so $$84 \times 33 = 2772$$.

Count = $$84 - 31 + 1 = 54$$

Result:

$$|A \cup B| = 600 + 164 - 54 = 710$$

The answer is $$\boxed{710}$$.

We need to count 3-digit numbers (100-999) divisible by 2 or 3 but not by 7.

First, we count numbers divisible by 2 or 3.

Divisible by 2: $$\lfloor 999/2 \rfloor - \lfloor 99/2 \rfloor = 499 - 49 = 450$$

Divisible by 3: $$\lfloor 999/3 \rfloor - \lfloor 99/3 \rfloor = 333 - 33 = 300$$

Divisible by 6 (both 2 and 3): $$\lfloor 999/6 \rfloor - \lfloor 99/6 \rfloor = 166 - 16 = 150$$

Divisible by 2 or 3: $$450 + 300 - 150 = 600$$

Next, we subtract those also divisible by 7.

We need numbers divisible by (2 or 3) AND 7 = divisible by 14 or 21.

Divisible by 14: $$\lfloor 999/14 \rfloor - \lfloor 99/14 \rfloor = 71 - 7 = 64$$

Divisible by 21: $$\lfloor 999/21 \rfloor - \lfloor 99/21 \rfloor = 47 - 4 = 43$$

Divisible by 42 (LCM of 14 and 21): $$\lfloor 999/42 \rfloor - \lfloor 99/42 \rfloor = 23 - 2 = 21$$

Divisible by 14 or 21: $$64 + 43 - 21 = 86$$

Final answer:

$$ 600 - 86 = 514 $$

Therefore, the number of 3-digit numbers is 514.

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 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 ($$3^2 \equiv 4 \pmod{5}$$).

The permitted digit set is $$\{0,\,2,\,3,\,4,\,6,\,7\}$$ (six digits).
Every required integer is 4-digit and must lie in the closed interval $$[2022,\,4482]$$.

Denote the thousands, hundreds, tens and units digits by $$T,H,E,U$$ respectively.
Because $$2022\le T\,H\,E\,U\le 4482$$, the thousands digit $$T$$ can take only the values $$2,3,4$$ (digits $$0,6,7$$ are either too small or would exceed $$4482$$).

The smallest admissible number is $$2022$$, so the last three-digit part $$H\,E\,U$$ must satisfy $$H\,E\,U\ge 022$$.

Total 3-digit strings possible with the allowed digits: $$6^3 = 216$$.

We must exclude those strings for which $$H\,E\,U\lt 022$$, i.e. $$000$$ to $$021$$. Among these first 22 numbers, only the following possess digits solely from the permitted set: $$000,002,003,004,006,007,020$$  —  7 strings in all.

Hence valid numbers when $$T=2$$: $$216-7 = 209$$.

All numbers $$3000$$-$$3999$$ automatically lie inside the interval. Each of $$H,E,U$$ can again be any of the 6 allowed digits, giving $$6^3 = 216$$ valid numbers.

Now the highest permissible integer is $$4482$$, so the last three-digit part must satisfy $$H\,E\,U\le 482$$.

Total possible 3-digit strings: $$6^3 = 216$$.
We count directly how many of these are $$\le 482$$.

The hundreds digit $$H$$ can be $$0,2,3,4$$ (digits $$6,7$$ would already yield $$\ge 600$$).

For $$H=0,2,3$$ every choice of $$E,U$$ works (because the number is $$\le 399$$). That yields $$3\times 6\times 6 = 108$$ numbers.

For $$H=4$$ we need $$4E U \le 482 \Longrightarrow 10E+U\le 82$$.
For each allowed tens digit $$E=0,2,3,4,6,7$$ the inequality $$U\le 82-10E$$ is satisfied by all six allowed units digits (each is at most $$7$$). Thus $$6\times 6 = 36$$ additional numbers.

Total for $$T=4$$: $$108+36 = 144$$.

Adding the three cases:

$$209 + 216 + 144 = 569$$

Therefore, the number of 4-digit integers required is 569.

We have $$S = \{4, 6, 9\}$$ and $$T = \{9, 10, 11, \ldots, 1000\}$$. The set $$A$$ consists of all numbers that can be represented as $$a_1 + a_2 + \cdots + a_k$$ where each $$a_i \in \{4, 6, 9\}$$ and $$k \in \mathbb{N}$$. We need to find the sum of all elements in $$T \setminus A$$, i.e., elements of $$T$$ that cannot be expressed as such sums.

We note that every element of $$A$$ is a non-negative integer combination of 4, 6, and 9 (with at least one term). The elements of $$A$$ starting from the smallest are: 4, 6, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18, ... We need to find which numbers from 9 onwards are NOT in $$A$$.

We systematically check small values. A number $$n$$ is in $$A$$ if $$n = 4a + 6b + 9c$$ for non-negative integers $$a, b, c$$ with $$a + b + c \geq 1$$:

$$n = 4$$: $$4 = 4(1)$$. Yes. $$n = 6$$: $$6 = 6(1)$$. Yes. $$n = 8$$: $$8 = 4(2)$$. Yes. $$n = 9$$: $$9 = 9(1)$$. Yes. $$n = 10$$: $$10 = 4+6$$. Yes. $$n = 12$$: $$12 = 4(3)$$ or $$6(2)$$. Yes. $$n = 13$$: $$13 = 4+9$$. Yes. $$n = 14$$: $$14 = 4+4+6$$. Yes. $$n = 15$$: $$15 = 6+9$$. Yes. $$n = 16$$: $$16 = 4(4)$$. Yes. $$n = 17$$: $$17 = 4+4+9$$. Yes. $$n = 18$$: $$18 = 9(2)$$ or $$4(3)+6$$. Yes.

Now, $$n = 19$$: $$19 = 4+6+9$$. Yes. $$n = 20$$: $$20 = 4(5)$$. Yes. For $$n = 21$$: $$21 = 12+9 = 4(3)+9$$. Yes. $$n = 22$$: $$22 = 4+9+9 = 4+18$$. Yes. $$n = 23$$: $$23 = 4+4+6+9 = 14+9$$. Yes.

We observe that all integers $$\geq 24$$ can be written as sums: if $$n \equiv 0 \pmod{4}$$, use $$4$$'s; if $$n \equiv 1 \pmod{4}$$, use $$9 + 4$$'s (since $$n \geq 25$$ gives enough room, and $$n = 9$$ works, $$n = 13$$ works); if $$n \equiv 2 \pmod{4}$$, use $$6 + 4$$'s; if $$n \equiv 3 \pmod{4}$$, use $$9 + 6 + 4$$'s.

More precisely, for $$n \geq 24$$: $$n \equiv 0 \pmod{2}$$ can always be written as $$4a + 6b$$; $$n$$ odd and $$n \geq 9$$ can be written as $$9 + \text{even number} \geq 0$$. Since $$n - 9 \geq 15$$ for $$n \geq 24$$, and $$n - 9$$ is even, and any even number $$\geq 4$$ can be written as $$4a + 6b$$ (and even numbers $$\geq 4$$: 4, 6, 8, 10, ... all work). Actually even number 0 also works (take nothing extra). So for odd $$n \geq 9$$: $$n - 9$$ is even and $$\geq 0$$, and we need $$n - 9 \neq 2$$ (since 2 cannot be made from 4s and 6s). $$n - 9 = 2 \Rightarrow n = 11$$. So $$n = 11$$ is the concern.

Let us check $$n = 11$$: Can we write $$11 = 4a + 6b + 9c$$? With $$c = 1$$: $$2 = 4a + 6b$$, no non-negative solution. With $$c = 0$$: $$11 = 4a + 6b$$, impossible (parity: $$4a + 6b$$ is even, 11 is odd). So $$n = 11 \notin A$$.

We also need to check numbers less than 24 more carefully. From the set $$T = \{9, 10, \ldots, 1000\}$$, which values from 9 to 23 are NOT in $$A$$?

$$n = 9$$: Yes (checked). $$n = 10$$: Yes. $$n = 11$$: No (shown above). $$n = 12$$: Yes. $$n = 13$$: Yes. $$n = 14$$: Yes. $$n = 15$$: Yes. $$n = 16$$: Yes. $$n = 17$$: Yes. $$n = 18$$: Yes. $$n = 19$$: Yes. $$n = 20$$: Yes. $$n = 21$$: Yes. $$n = 22$$: Yes. $$n = 23$$: Yes.

For $$n \geq 24$$, every number is representable as argued. So the only element of $$T$$ not in $$A$$ is $$n = 11$$.

Therefore, the sum of all elements in $$T \setminus A$$ is $$\boxed{11}$$.

Hence, the correct answer is 11.

We need to find the total number of 3-digit numbers whose GCD with 36 is 2.

First, factorizing 36 gives $$36 = 2^2 \times 3^2$$.

Next, for $$\gcd(n, 36) = 2$$, we require $$n$$ to be even but not divisible by 4 (so that the highest power of 2 dividing the GCD is exactly $$2^1$$), and also not divisible by 3. In other words, $$n = 2m$$ where $$m$$ is odd and not divisible by 3.

Now, considering 3-digit numbers with $$100 \le 2m \le 999$$, we get $$50 \le m \le 499$$.

Since $$m$$ must be odd and not divisible by 3, we first count the odd integers from 51 to 499, which number $$\frac{499 - 51}{2} + 1 = 225$$.

Subsequently, we remove the odd multiples of 3 in the same range. These are numbers of the form $$3k$$ where $$k$$ is odd and $$51 \le 3k \le 499$$. The smallest such number is 51 ($$= 3 \times 17$$) and the largest is 495 ($$= 3 \times 165$$), giving the sequence $$3 \times 17, 3 \times 19, \dots, 3 \times 165$$, which has $$\frac{165 - 17}{2} + 1 = 75$$ terms.

Therefore, the total count is $$225 - 75 = 150$$, so the answer is $$150$$.

We need the sum of all $$\alpha \in \{1, 2, \ldots, 100\}$$ such that $$\gcd(\alpha, 24) = 1$$.

$$24 = 2^3 \times 3$$. So $$\gcd(\alpha, 24) = 1$$ means $$\alpha$$ is not divisible by 2 or 3.

We use inclusion-exclusion. The sum of all integers from 1 to 100 is $$\frac{100 \times 101}{2} = 5050$$.

Sum of multiples of 2 from 1 to 100: $$2 + 4 + \ldots + 100 = 2 \times \frac{50 \times 51}{2} = 2550$$.

Sum of multiples of 3 from 1 to 100: $$3 + 6 + \ldots + 99 = 3 \times \frac{33 \times 34}{2} = 3 \times 561 = 1683$$.

Sum of multiples of 6 from 1 to 100: $$6 + 12 + \ldots + 96 = 6 \times \frac{16 \times 17}{2} = 6 \times 136 = 816$$.

By inclusion-exclusion, the sum of integers divisible by 2 or 3:

$$S_{2 \cup 3} = 2550 + 1683 - 816 = 3417$$

The sum of integers coprime to 24:

$$S = 5050 - 3417 = 1633$$

The correct answer is $$1633$$.

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 to find the total number of positive integral solutions $$(x, y, z)$$ such that $$xyz = 24$$.

First, we find the prime factorization: $$24 = 2^3 \times 3$$. The divisors of 24 are $$1, 2, 3, 4, 6, 8, 12, 24$$.

For each value of $$x$$ that divides 24, we count the number of ordered pairs $$(y, z)$$ such that $$yz = \frac{24}{x}$$. This count equals the number of divisors of $$\frac{24}{x}$$, denoted $$d\!\left(\frac{24}{x}\right)$$.

Computing for each divisor: when $$x = 1$$, $$yz = 24$$ which has $$d(24) = 8$$ solutions; when $$x = 2$$, $$yz = 12$$ which has $$d(12) = 6$$ solutions; when $$x = 3$$, $$yz = 8$$ which has $$d(8) = 4$$ solutions; when $$x = 4$$, $$yz = 6$$ which has $$d(6) = 4$$ solutions; when $$x = 6$$, $$yz = 4$$ which has $$d(4) = 3$$ solutions; when $$x = 8$$, $$yz = 3$$ which has $$d(3) = 2$$ solutions; when $$x = 12$$, $$yz = 2$$ which has $$d(2) = 2$$ solutions; when $$x = 24$$, $$yz = 1$$ which has $$d(1) = 1$$ solution.

The total number of ordered triples is $$8 + 6 + 4 + 4 + 3 + 2 + 2 + 1 = 30$$.

Therefore, the total number of positive integral solutions is $$\boxed{30}$$.

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 need to find the number of two-digit numbers $$n$$ such that $$3^n + 7^n$$ is a multiple of 10. For this, we need $$3^n + 7^n \equiv 0 \pmod{10}$$.

We examine the last digits of powers of 3 and 7. The cycle of last digits of $$3^n$$ is: $$3, 9, 7, 1$$ (period 4). The cycle of last digits of $$7^n$$ is: $$7, 9, 3, 1$$ (period 4).

Adding the last digits for each $$n \bmod 4$$: when $$n \equiv 1 \pmod{4}$$, last digits sum to $$3 + 7 = 10$$; when $$n \equiv 2 \pmod{4}$$, last digits sum to $$9 + 9 = 18$$; when $$n \equiv 3 \pmod{4}$$, last digits sum to $$7 + 3 = 10$$; when $$n \equiv 0 \pmod{4}$$, last digits sum to $$1 + 1 = 2$$.

So $$3^n + 7^n$$ is divisible by 10 when the last digit sum is 10 or 20, etc. From above, this happens when $$n \equiv 1$$ or $$3 \pmod{4}$$, i.e., when $$n$$ is odd.

The two-digit numbers range from 10 to 99. The odd two-digit numbers are 11, 13, 15, ..., 99. Their count is $$\frac{99 - 11}{2} + 1 = 45$$.

Therefore, the total number of such two-digit numbers is $$45$$.

We need the sum of all $$n \in \{1, 2, \ldots, 100\}$$ such that $$\gcd(n, 2040) = 1$$. First, factorize $$2040 = 2^3 \times 3 \times 5 \times 17$$. So $$n$$ must not be divisible by 2, 3, 5, or 17.

We use inclusion-exclusion on the sum. Let $$S_d$$ denote the sum of all multiples of $$d$$ in $$\{1, \ldots, 100\}$$. If there are $$k = \lfloor 100/d \rfloor$$ such multiples, then $$S_d = d \cdot \frac{k(k+1)}{2}$$.

The sum we want is $$S_{\text{total}} - (S_2 + S_3 + S_5 + S_{17}) + (S_6 + S_{10} + S_{34} + S_{15} + S_{51} + S_{85}) - (S_{30} + S_{102} + S_{170} + S_{255}) + S_{510}$$, where terms with $$d > 100$$ contribute zero.

Computing each: the total sum is $$\frac{100 \times 101}{2} = 5050$$. For $$S_2$$: $$k = 50$$, $$S_2 = 2 \times 1275 = 2550$$. For $$S_3$$: $$k = 33$$, $$S_3 = 3 \times 561 = 1683$$. For $$S_5$$: $$k = 20$$, $$S_5 = 5 \times 210 = 1050$$. For $$S_{17}$$: $$k = 5$$, $$S_{17} = 17 \times 15 = 255$$.

For the pairwise terms: $$S_6$$: $$k = 16$$, $$S_6 = 6 \times 136 = 816$$. $$S_{10}$$: $$k = 10$$, $$S_{10} = 10 \times 55 = 550$$. $$S_{34}$$: $$k = 2$$, $$S_{34} = 34 \times 3 = 102$$. $$S_{15}$$: $$k = 6$$, $$S_{15} = 15 \times 21 = 315$$. $$S_{51}$$: $$k = 1$$, $$S_{51} = 51$$. $$S_{85}$$: $$k = 1$$, $$S_{85} = 85$$.

For the triple terms: $$S_{30}$$: $$k = 3$$, $$S_{30} = 30 \times 6 = 180$$. $$S_{102}$$, $$S_{170}$$, $$S_{255}$$ all have $$k = 0$$, contributing nothing. The quadruple term $$S_{510} = 0$$ as well.

Putting it all together: $$5050 - (2550 + 1683 + 1050 + 255) + (816 + 550 + 102 + 315 + 51 + 85) - 180 = 5050 - 5538 + 1919 - 180 = 1251$$.

We have $$A = \{n \in \mathbb{N} : n \text{ is a 3-digit number}\} = \{100, 101, \ldots, 999\}$$, $$B = \{9k + 2 : k \in \mathbb{N}\}$$, and $$C = \{9k + l : k \in \mathbb{N}\}$$ where $$0 < l < 9$$.

The set $$B$$ consists of all natural numbers that leave remainder $$2$$ when divided by $$9$$. Similarly, $$C$$ consists of all natural numbers that leave remainder $$l$$ when divided by $$9$$.

The 3-digit numbers with remainder $$r$$ modulo $$9$$ (for $$1 \leq r \leq 8$$) form an arithmetic progression with first term $$99 + r$$, last term $$990 + r$$, and common difference $$9$$. The number of such terms is $$\frac{(990 + r) - (99 + r)}{9} + 1 = 100$$.

The sum of the 3-digit numbers with remainder $$r$$ is $$\frac{100}{2} \times (99 + r + 990 + r) = 50(1089 + 2r)$$.

For $$B$$, the remainder is $$r = 2$$, so the sum of 3-digit numbers in $$B$$ is $$50(1089 + 4) = 50 \times 1093 = 54650$$.

If $$l \neq 2$$, then $$B$$ and $$C$$ pick out different residue classes, and $$A \cap (B \cup C)$$ is the union of two disjoint sets. The total sum is $$54650 + 50(1089 + 2l)$$.

We are given that this sum equals $$274 \times 400 = 109600$$. So $$54650 + 50(1089 + 2l) = 109600$$.

Simplifying, $$54650 + 54450 + 100l = 109600$$, which gives $$109100 + 100l = 109600$$, so $$100l = 500$$ and $$l = 5$$.

Hence, the answer is $$5$$.

We have to count those positive divisors that can be written in the form $$4n+1$$ for the number $$ (10)^{10}\,(11)^{11}\,(13)^{13}. $$ First we write every factor in terms of its prime constituents:

$$10 = 2\cdot5,$$ so $$ (10)^{10}=2^{10}\,5^{10}.$$ Hence the complete prime‐factorisation of the given number is

$$ N \;=\;2^{10}\,5^{10}\,11^{11}\,13^{13}. $$

Any divisor $$d$$ of $$N$$ therefore has the general form

$$ d \;=\;2^{a}\,5^{b}\,11^{c}\,13^{d_1}, $$

where the exponents are restricted by the powers present in $$N$$:

$$ 0\le a\le 10,\quad 0\le b\le 10,\quad 0\le c\le 11,\quad 0\le d_1\le 13. $$

To ensure that $$d$$ is of the form $$4n+1$$ we must have $$d\equiv1\pmod4.$$ We therefore study each prime factor modulo $$4$$:

$$ 2\equiv2\pmod4,\qquad5\equiv1\pmod4,\qquad11\equiv3\pmod4,\qquad13\equiv1\pmod4. $$

Because $$5\equiv1\pmod4$$ and $$13\equiv1\pmod4,$$ every power of $$5$$ or $$13$$ is again congruent to $$1$$ modulo $$4$$:

$$ 5^{b}\equiv1\pmod4,\qquad13^{d_1}\equiv1\pmod4\quad\text{for all }b,d_1. $$

Only the powers of $$2$$ and $$11$$ can change the final remainder. We therefore examine them separately.

Behaviour of $$2^{a}\pmod4$$

$$ 2^{0}=1,\quad2^{1}=2,\quad2^{2}=4\equiv0\pmod4, $$ and for any $$a\ge2$$ we still have $$2^{a}\equiv0\pmod4.$$ Hence

$$ 2^{a}\equiv \begin{cases} 1,&a=0,\\ 2,&a=1,\\ 0,&a\ge2. \end{cases} $$

If $$a\ge2$$ the factor $$2^{a}$$ already forces the whole product to be $$0\pmod4,$$ so such divisors can never be congruent to $$1\pmod4.$$ Therefore we must restrict ourselves to

$$ a=0\quad\text{or}\quad a=1. $$

Behaviour of $$11^{c}\pmod4$$

Since $$11\equiv3\pmod4,$$ we use the basic property $$3^{2}\equiv9\equiv1\pmod4.$$ Thus

$$ 11^{c}\equiv3^{c}\equiv \begin{cases} 1,&c\text{ even},\\ 3,&c\text{ odd}. \end{cases} $$

Combining the two factors

The total remainder contributed by $$2^{a}11^{c}$$ is

$$ (2^{a}\bmod4)\times(3^{c}\bmod4)\pmod4. $$

We now test the only possible values of $$a$$(namely $$0$$ and $$1$$):

• If $$a=0,$$ then $$2^{a}\equiv1,$$ so we need $$ 1\times3^{c}\equiv1\pmod4, $$ which is true exactly when $$3^{c}\equiv1,$$ i.e. when $$c$$ is even.

• If $$a=1,$$ then $$2^{a}\equiv2,$$ giving $$ 2\times3^{c}\pmod4\in\{2,\,2\}\quad(\text{because }3^{c}\equiv1\text{ or }3), $$ both of which are $$2\pmod4,$$ never $$1.$$ Hence no divisor with $$a=1$$ can satisfy the condition.

Consequently the required divisors must have

$$ a=0,\qquad c\text{ even}. $$

Counting the allowable choices

• For $$a$$ there is only one choice: $$a=0.$$
• For $$c$$ (ranging from $$0$$ to $$11$$) the even values are $$0,2,4,6,8,10,$$ a total of $$6$$ choices.
• For $$b$$ any value from $$0$$ to $$10$$ is permitted, giving $$10-0+1=11$$ choices.
• For $$d_1$$ any value from $$0$$ to $$13$$ is allowed, giving $$13-0+1=14$$ choices.

Multiplying the independent counts we obtain the total number of divisors congruent to $$1\pmod4$$:

$$ 1\times6\times11\times14 \;=\; 6\times154 \;=\; 924. $$

So, the answer is $$924$$.

We need to count 4-digit numbers $$n$$ (from 1000 to 9999) such that $$\gcd(n, 18) = 3$$. Since $$18 = 2 \cdot 3^2$$, the condition $$\gcd(n, 18) = 3$$ requires: (i) $$3 \mid n$$ (so $$\gcd$$ includes at least one factor of 3), (ii) $$9 \nmid n$$ (otherwise $$\gcd$$ would include $$9$$), and (iii) $$2 \nmid n$$ (otherwise $$\gcd$$ would include a factor of 2).

So we need $$n$$ to be odd, divisible by 3, but not divisible by 9.

The number of odd 4-digit numbers is $$\frac{9999 - 1001}{2} + 1 = 4500$$. The number of odd multiples of 3 in this range equals $$\frac{4500}{3} = 1500$$ (since among any three consecutive odd numbers, exactly one is divisible by 3).

The number of odd multiples of 9 in $$[1000, 9999]$$: The multiples of 9 in this range are $$1008, 1017, \ldots, 9999$$, totalling $$\frac{9999 - 1008}{9} + 1 = 1000$$. Exactly half of these are odd (since consecutive multiples of 9 alternate parity as 9 is odd), giving 500 odd multiples of 9.

Therefore, the count of 4-digit numbers with $$\gcd(n, 18) = 3$$ is $$1500 - 500 = 1000$$.

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.