Unit 6 - Practice Quiz

Modular arithmetic finds the remainder of a division. When 17 is divided by 5, the quotient is 3 and the remainder is 2. Therefore, .

The statement " divides " means that is a multiple of . This can be expressed algebraically as , where is an integer.

A prime number is a natural number greater than 1 that has no positive divisors other than 1 and itself. 23 is only divisible by 1 and 23. 1 is a special case (neither prime nor composite), 9 is divisible by 3, and 15 is divisible by 3 and 5.

The Fundamental Theorem of Arithmetic guarantees that every integer greater than 1 is either a prime itself or can be factored into a product of prime numbers, and this factorization is unique (apart from the order of the factors).

The divisors of 12 are {1, 2, 3, 4, 6, 12}. The divisors of 18 are {1, 2, 3, 6, 9, 18}. The greatest divisor that they have in common is 6.

The multiples of 10 are {10, 20, 30, 40, ...}. The multiples of 15 are {15, 30, 45, ...}. The smallest multiple that they have in common is 30.

The primary purpose of the Euclidean algorithm is to find the greatest common divisor (GCD) of two integers by using repeated division with remainder.

The first step is to express 70 in terms of 21: . The remainder is 7. The next step would be to find .

Bezout's lemma states that the greatest common divisor of and is the smallest positive integer that is a linear combination of and .

A linear congruence is an equation of the form , where is the variable we are trying to solve for. It is 'linear' because the variable is raised to the first power.

We can test the values. If , then . Since , we have . Therefore, is a solution.

An integer has a multiplicative inverse modulo if and only if and are relatively prime (or coprime), which means their greatest common divisor is 1.

We are looking for an integer such that . Testing the options, we find that . Since , we have . Thus, 4 is the inverse.

The Chinese Remainder Theorem provides a method to find a unique solution for a system of simultaneous linear congruences, given that the moduli are pairwise coprime.

The Caesar cipher is a substitution cipher where each letter is shifted a certain number of places down the alphabet. Shifting 'B' by 3 positions results in 'E' (B -> C -> D -> E).

For the English alphabet, there are only 25 possible keys (shifts). An attacker can simply try all 25 possibilities until the decrypted message is readable, which is a simple brute-force attack.

The Caesar cipher is a special case of the Affine cipher where . The encryption function for a Caesar cipher with key is , which matches the affine form with and .

Fermat's Little Theorem states that if is a prime, then for any integer not divisible by , the congruence holds true.

The definition of congruence is that divides the difference . This is equivalent to saying that is a multiple of .

By definition, two integers are relatively prime if they share no common divisors other than 1. Therefore, their greatest common divisor must be 1.

We can use properties of modular arithmetic. First, find the remainders of each number when divided by 11:

• 
  Now substitute these into the expression:
  
  
  
  .
  The remainder is 7.

The steps of the Euclidean algorithm are:

1. (Remainder is 112)
2. (Remainder is 36)
3. (Remainder is 4)
4. (Remainder is 0)
   The sequence of remainders is 112, 36, 4, and finally 0. The GCD is the last non-zero remainder, which is 4.

Bezout's lemma states that the equation has integer solutions if and only if is a multiple of .
First, we find the gcd of 56 and 72 using the Euclidean algorithm:



The gcd is 8. Therefore, the equation has solutions if and only if is a multiple of 8. Among the options, only 40 is a multiple of 8.

A linear congruence has solutions if and only if divides . If it does, there are exactly incongruent solutions.
Here, .
First, find . The divisors of 12 are {1, 2, 3, 4, 6, 12}. The divisors of 30 are {1, 2, 3, 5, 6, 10, 15, 30}. The greatest common divisor is 6. So, .
Next, check if divides . Does 6 divide 18? Yes, . Since the condition is met, there are exactly incongruent solutions.

We need to find an integer such that . We use the Extended Euclidean Algorithm to find integers and such that .




Working backwards:




So, . Taking this equation modulo 45, we get . The inverse is 8.

From the first congruence, must be of the form for some integer .
Substitute this into the second congruence:


To solve for , we need the inverse of 3 modulo 5. Since , the inverse is 2.
Multiply both sides by 2:



So, must be of the form for some integer .
The smallest non-negative value for is 2 (when ).
Substitute this value of back into the expression for : .
To check: and . The smallest positive integer is 8.

Here, and . So, .
We need to express the exponent 1004 in terms of 12. We divide 1004 by 12:
.
So, .
Now we apply the modular arithmetic:

.
Let's calculate :


.
The remainder is 9.

The decryption function for an affine cipher is .
Here, . First, we need to find the inverse of modulo 26. We need to solve .
By inspection, . , so .
Thus, .
Now, substitute into the decryption formula:


.
We need to simplify . . So, .
Let's re-calculate .
.

We use the fundamental relationship between two positive integers and and their gcd and lcm: .
Given values are: , , and .
We need to find . Plugging the values into the formula:


.
The other number is 36. We can verify: gcd(60, 36) = 12 and lcm(60, 36) = 180.

To find the number of positive divisors of an integer, we first find its prime factorization.



Combining these, we get the prime factorization of 720:
.
If the prime factorization of a number is , the number of divisors is given by the product .
For 720, the exponents are 4, 2, and 1. So, the number of divisors is:
.

First, find the gcd of 87 and 54:





. The gcd is 3.
Now, work backwards to express 3 in the form :




.

First, we must find the key (the shift value). We use the numerical representation A=0, B=1, ..., Z=25. The letter 'F' is the 5th letter (index 5). The letter 'A' is the 0th letter (index 0). The encryption formula is . We have , which means the key . To decrypt, we use the formula . So we need to shift each ciphertext letter back by 5 positions. - Y (24) -> -> T. Whoops, this isn't matching APPLE. Let me check. Ciphertext F (5) is plaintext A (0). So . The shift is +5. To decrypt YJQQF, we shift back 5. Y (24) -> -> T J (9) -> -> E Q (16) -> -> L Q (16) -> -> L F (5) -> -> A Decryption is TELIA. My question is wrong. Let's make it work for APPLE. Plaintext APPLE: A(0), P(15), P(15), L(11), E(4). If key is k=24 (or -2), A(0) -> 24 (Y). P(15) -> (N). Doesn't work. Let's work from Ciphertext to Plaintext. Ciphertext YJQQF. Let's say plaintext is VILLA. V(21) -> I(8) -> L(11) -> L(11) -> A(0). Key must be constant. Let's try again. Ciphertext 'YJQQF'. Plaintext 'APPLE'. A (0) -> Y (24). Key k = 24 or -2. P (15) -> (N). Ciphertext would be YNN.... Let's re-do the question from scratch. Plaintext: HELLO. Key k=10. H(7) -> 17 (R). E(4) -> 14 (O). L(11) -> 21 (V). L(11) -> 21 (V). O(14) -> 24 (Y). Ciphertext: ROVVY. Question: Ciphertext is ROVVY. V decrypts to L. What is the plaintext? V(21) decrypts to L(11). . . k=10. Decrypt ROVVY with k=10. R(17) -> 7 (H). O(14) -> 4 (E). V(21) -> 11 (L). V(21) -> 11 (L). Y(24) -> 14 (O). HELLO. This works. Question: The ciphertext "ROVVY" was encrypted using a Caesar cipher. If the letter 'V' in the ciphertext corresponds to the plaintext letter 'L', what is the original message? Options: HELLO, WORLD, GREEN, JELLO. Correct: HELLO. Explanation: Let A=0, ..., Z=25. The decryption is . We are given that ciphertext 'V' (21) is plaintext 'L' (11). . The shift key is 10. Now we decrypt the entire message "ROVVY" by shifting each letter back by 10. - R(17): H - O(14): E - V(21): L - V(21): L - Y(24): O The original message is HELLO.

Finding the last digit of a number is equivalent to finding the number modulo 10. We need to compute . Let's look at the pattern of powers of 3 modulo 10:





The pattern of the last digits is 3, 9, 7, 1, and it repeats every 4 powers. The length of the cycle is 4.
To find the last digit of , we need to find the exponent modulo 4.
with a remainder of 3. So, .
This means that will have the same last digit as . .
The last digit is 7.

A multiplicative inverse of 'a' modulo 'm' exists if and only if 'a' and 'm' are relatively prime, which means their greatest common divisor is 1 (i.e., ).
In this case, . The prime factors of 34 are 2 and 17. An inverse for 'a' will not exist if 'a' shares a common factor with 34, meaning 'a' is a multiple of 2 or 17.
We check the gcd for each option:

• (15 is not divisible by 2 or 17).
• (19 is not divisible by 2 or 17).
• (25 is not divisible by 2 or 17).
• (Both are even numbers).
  Since , a multiplicative inverse for 28 modulo 34 does not exist.

Let's perform the Euclidean Algorithm for gcd(1001, 343):
Step 1: .
. So, the first quotient and first remainder .
Step 2: .
. So, the second quotient and second remainder .
Step 3: .
.
Step 4: .
.
The question asks for the second quotient, , which is 1.

We need to solve for in .
Since 23 is a small prime, we can try to simplify the coefficient 19.
. So the congruence becomes:
.
To remove the negative, we can multiply by -1:
.
And . So:
.
Since , we can divide both sides by 4.
.
.
Now, multiply both sides of the original congruence by the inverse, 3:



.
The set of solutions is . The smallest positive integer solution is 7.

Wilson's Theorem states that for a prime , we have . For , this gives us .
We want to find . We can write as .
So, .
In modulo 17, the number 16 is equivalent to -1. So we can substitute this into the congruence:
.
To solve for , we can multiply both sides by -1:


.
The value is 1.

This problem can be translated into a system of linear congruences. Let be the number of students.

2. 
   From the second congruence, we know that can be written as for some integer .
   Substitute this expression into the first congruence:
   
   Subtract 1 from both sides:
   
   Since , we have:
   
   To solve for , we need the inverse of 2 modulo 5. The inverse is 3, since .
   Multiply both sides by 3:
   
   
   
   So, must be of the form . The smallest non-negative value for is 4 (when ).
   Now substitute this value of back into the equation for :
   .
   The smallest possible number of students is 29.

We are given two plaintext-ciphertext pairs which we can use to form a system of linear congruences.

1. For B(1) F(5):
2. For D(3) L(11):
   Subtract the first equation from the second:
   
   
   This congruence means is a multiple of 26. Dividing by 2, we get is a multiple of 13. So, . This means could be 3 or 16. We must check which one works. An additional condition for an affine cipher is that . Since , cannot be 16. Therefore, .
   Now substitute back into the first equation:
   
   .
   The encryption key is .

The problem asks for the first time the two bells will toll together again after the start. This is a classic application of the least common multiple (lcm).
We need to find the lcm of 42 and 56.
First, find the prime factorization of each number:


To find the lcm, we take the highest power of each prime factor present in the factorizations and multiply them together.
The prime factors are 2, 3, and 7.
Highest power of 2 is .
Highest power of 3 is .
Highest power of 7 is .
.
They will next toll together after 168 seconds.

The moduli are not pairwise coprime, so we must first break the system down into congruences with prime power moduli.
, , .

Now, we check for consistency. The condition implies is odd, which is consistent with . The condition implies , so , which is consistent. The conditions are consistent.

The reduced, equivalent system is:

The moduli are now pairwise coprime (4, 5, 9). We solve this using the Chinese Remainder Theorem. The solution will be unique modulo .
From , we have .
Substitute into the second congruence: . Multiplying by $4$ (the inverse of ), we get . So .
Substitute back into : .
Substitute this into the first congruence: . So .
Finally, substitute back into : .
So, .
We need the smallest solution greater than 2000. We want . The smallest integer is 11.
For , .

Wait, I made a mistake in the calculation. Let me re-check. . . . . So the exponent is . The result is . What if the exponent mod 16 was not 1? Let's use . . We need . . . . . Since , . So the exponent is . Then . This is not very interesting. Let's try to get a more complex exponent. Exponent is . . Since , by Euler's totient theorem, . . . . So . Exponent is . Still gives 7. Let's go back to the original question and re-verify my steps. . Exponent mod 16. . . Need . . . Exponent is . . Perhaps I should make the base and modulus different. Let's compute . Exponent mod 22. . . We need . . Since the exponent is large, is divisible by 2. Also, we need to consider modulo 11. . . . We need . By FLT, . . . So the exponent satisfies: and . . . Multiply by $6$ (inv of 2): . . . So . The original problem becomes . . . . . . This is a good hard question. Let me re-write the question and solution. Question: What is the remainder when is divided by $19$? Exponent mod 18. . . So . We need . We need . By Euler's totient theorem, . . So . . So . The exponent is of the form . The problem becomes . We need to compute . . . . . . . So . The final answer is 11. This seems correct and sufficiently hard. The options: 11, 1, 7, 13. The correct option is 11. Explanation needs to be re-written carefully.

The linear congruence has a solution for if and only if has a multiplicative inverse modulo 600. This is true if and only if .
The question asks for the number of integers in the range such that . This is almost the definition of Euler's totient function, .

counts the number of positive integers up to a given integer that are relatively prime to . The range is typically .
So, we need to calculate .

First, find the prime factorization of 600:
.

Now, use the formula for Euler's totient function:



.

This counts the numbers in the range with . The question asks for the range . Since , the number 600 is not counted by . Therefore, the count for the range is the same as the count for , which is .

We need to solve the congruence .
First, we find the greatest common divisor .
Using the Euclidean algorithm:




So, .

A solution exists if and only if divides , which is $102$. Since , a solution exists. The congruence has exactly incongruent solutions modulo 120.

To find the solutions, we first divide the entire congruence by :

Now, we need to find the inverse of 13 modulo 20. We can see that , so .
Alternatively, use the extended Euclidean algorithm:



. So . The inverse is .

Multiply the reduced congruence by the inverse (17):
.
, so .
This is the principal solution, . The general solution is for some integer .

The 6 incongruent solutions modulo 120 are given by for .
.
Solutions are:

The sum of these solutions is . This is an arithmetic series with , first term , and last term .

According to a property that is a direct consequence of Bezout's Lemma, a linear Diophantine equation of the form has integer solutions for and if and only if divides .
Therefore, the set of all possible values for is the set of all integer multiples of .

We need to compute . We use the Euclidean Algorithm:

The greatest common divisor is .

Thus, the equation has integer solutions if and only if is a multiple of 7. The set of all possible integer values for the expression is the set of all integer multiples of 7.

The encryption function is . We are given two plaintext-ciphertext pairs:

1. i (x=8) encrypts to X (y=23):
2. f (x=5) encrypts to C (y=2):

We solve this system of linear congruences. Subtract the second equation from the first:

To solve for , we need the inverse of 3 modulo 26. We can see that . So, .
Multiply the congruence by 9:


Since , we have .

Now, substitute into the second original equation:



.

The encryption key is . The encryption function is .

To decrypt, we need the decryption function .
We need . We can test values or use the Extended Euclidean Algorithm. , so .
.
The decryption function is .

Now we decrypt the ciphertext PK (y=15, y=10):

1. Decrypt P (y=15):
   .
   eq 18a+b=173a = -3 = 233a \equiv 23 \pmod{26}$.$9 \cdot 3a \equiv 9 \cdot 23 \pmod{26} \implies a \equiv 207 \pmod{26}$. $207 = 7 \cdot 26 + 25a=25 \equiv -1$.$5(-1)+b=20 \implies -5+b=20 \implies b=25(-1, 25)$. $E(x) = -x+25a^{-1} = (-1)^{-1} = -1 = 25$.$D(y) = -1(y-25) = -y+25D(1) = -1+25=24D(24) = -24+25=18a+b=03a = -23 = 3a=1$.$b=-8a=-8=18(1, 18)E(x)=x+18$. $D(y)=y-18D(1)=1-18=-17=9D(24)=24-18=622a+b=1418a = -4 = 22$.$18a \equiv 22 \pmod{26}\gcd(18,26)=2$.$9a \equiv 11 \pmod{13}$.$9^{-1} \pmod{13}$. $9 \cdot 3 = 27 = 2 \cdot 13 + 1a \equiv 11 \cdot 3 = 33 \equiv 7 \pmod{13}a=7a=7+13=20a=74(7)+b = 28+b=2+b=18 \implies b=16(7,16)\gcd(7,26)=1a=204(20)+b=80+b=2+b=18 \implies b=16(20,16)$. $\gcd(20,26)=2
   eq 1(7,16)$.$a^{-1}=7^{-1}=15$. $D(y) = 15(y-16)D(15)=15(15-16)=15(-1)=-15=11D(10)=15(10-16)=15(-6)=-90 = -4(26)+14=14$ ('O').Decrypts to LO. This works and is complex enough. The plaintext we is high frequency.I'll replace the original question with this one.Question: we (22, 4) -> OS (14, 18). Decrypt PK (15, 10). Options: LO, HI, GO, BE.

This question analyzes the structure of Euclid's proof for the infinitude of primes.
Let's analyze the properties of .

When we divide by any of the first primes ( for ), we get a remainder of 1. For example, , where is the product of the other primes. This means that is not divisible by any of the primes .

By the Fundamental Theorem of Arithmetic, every integer greater than 1 is either a prime number itself or can be represented as a product of prime numbers.
Since , it must have a prime divisor.

Let's evaluate the options:

1. N is a prime number. This is not always true. A counterexample is for . The first 6 primes are 2, 3, 5, 7, 11, 13. . $30031$ is not prime, as .
2. N is a composite number. This is also not always true. For , primes are 2, 3. , which is prime. For , primes are 2, 3, 5. , which is prime.
3. N is divisible by a prime number greater than . We established that must have a prime divisor, and that this divisor cannot be any of . Therefore, any prime divisor of must be a prime that is not in the set of the first primes. This means it must be a prime greater than . This statement is always true.
4. N is divisible by . This is not necessarily true. In the counterexample for option 1, and . The prime factors of were 59 and 509, neither of which is 17. The prime factor is just some prime greater than , not necessarily the very next one.

This question concerns the cancellation law in modular arithmetic.
The given congruence is . By definition of modular congruence, this means that , or .

We are given that . This allows us to write and for some integers and where .

Substitute these into the divisibility relation:

We can divide both sides by (since as ):

Now, we have that divides the product . Since we know that and are relatively prime (i.e., ), we can apply Euclid's Lemma. Euclid's Lemma states that if an integer divides the product of two other integers and is coprime to one of them, it must divide the other.

Therefore, we can conclude that .

By definition of congruence, this means .
Substituting back , we get the conclusion:

This is the strongest possible conclusion. Option A is a special case where . Option C is weaker, because if , then it must be that since is a multiple of only if . The conclusion is not always implied, but the question asks for the strongest valid conclusion from the premise.

This question tests identities involving GCD and LCM for three variables.

1. : This identity is well-known for two variables (), but it is not true in general for three or more variables.

   • Counterexample: Let .
     • .
     • .
     • LHS = .
     • RHS = .
     • Since , the identity is false.

2. : This is the standard associative property used to compute the LCM of multiple numbers. It is always true. For the example above, , which is correct.

3. : This is the associative property for GCD, which is also always true. For the example above, , which is correct.

4. : This is a known, valid identity for three variables, sometimes called the symmetric identity. Let's check with the p-adic valuation (the exponent of a prime p in the factorization). Let be the exponent of in . We need to show .

   • .
   • The p-adic valuation of the RHS is .
   • By the Principle of Inclusion-Exclusion for min/max, this expression is indeed equal to . Thus, this identity is always true.

Therefore, the only relation that is not always true is the first one.

This problem is an application of Wilson's Theorem, which is closely related to Fermat's Little Theorem.
Wilson's Theorem states that for a prime number , we have .

Here, , which is a prime number. According to Wilson's Theorem:

We want to find . Let's expand until we reach :

Now, let's convert the terms to their equivalents modulo 101:

Substitute these into the congruence:

Let . We have the linear congruence . To solve for , we need to find the multiplicative inverse of 24 modulo 101. We use the Extended Euclidean Algorithm.

Now, express 1 in terms of 101 and 24:

From this, we see that , so .
The inverse of 24 is .

Now, multiply the congruence by 80:

Thus, the remainder of when divided by 101 is 21.

We need to solve . This is equivalent to .
We can use the Chinese Remainder Theorem (CRT) to solve this. First, find the prime factorization of the modulus: .

The congruence is equivalent to the system of congruences:

Let's simplify and find the number of solutions for each modulus:

1. Modulo 2: . The only solution is . (1 solution)
2. Modulo 3: . The solutions are and . (2 solutions)
3. Modulo 5: . The solutions are and . (2 solutions)
4. Modulo 13: . We need to check if 10 is a quadratic residue mod 13. We can check all squares: . So is a solution. The other solution is . (2 solutions)

By the Chinese Remainder Theorem, the total number of solutions to the system is the product of the number of solutions for each individual congruence.
Total number of solutions = (solutions for mod 2) (solutions for mod 3) (solutions for mod 5) (solutions for mod 13).
Total number of solutions = .
There are 8 incongruent solutions modulo 390.

This question is about the complexity of the Euclidean algorithm, specifically Lamé's theorem.

Lamé's theorem states that the number of steps required for the Euclidean algorithm for two positive integers is less than or equal to five times the number of decimal digits in the smaller number. This leads to the bound .

The worst-case scenario for the Euclidean algorithm occurs when the inputs are consecutive Fibonacci numbers. If we take and , the algorithm takes steps. The Fibonacci sequence is defined by . From this, we can derive a tighter bound in terms of logarithms.

Let's analyze the options:

• : This is a trivial upper bound, as in each step the remainder is strictly smaller than the previous one. It's not tight at all.
• : This is a valid, but not the tightest, logarithmic bound. It can be shown that the remainder is at least halved every two steps.
• : This is the common statement of Lamé's theorem. .
• : This is a very tight bound derived from the Fibonacci worst-case analysis. .

Comparing the logarithmic bounds, is generally tighter than . For example, if , the first bound is . The second bound is . The bound is known to be a more precise formulation of Lamé's theorem based on the connection to Fibonacci numbers, making it the tightest bound among the choices.

We need to test each option using Korselt's criterion.

• A) 341: 1. Prime factorization: . It is composite. 2. Is it square-free? Yes, the prime factors appear with power 1. 3. Check the divisibility condition: . For , we check if , i.e., . Yes, . For , we check if , i.e., . No, 340 is not divisible by 30. 4. Since the condition fails for , 341 is not a Carmichael number. (It is a Fermat pseudoprime to base 2, as ). B) 561: 1. Prime factorization: . It is composite. 2. Is it square-free? Yes. 3. Check the divisibility condition: . For : . Does ? Yes. For : . Does ? Yes. For : . Does ? Yes, . 4. All conditions of Korselt's criterion are met. Therefore, 561 is a Carmichael number. C) 729: 1. Prime factorization: . It is composite. 2. Is it square-free? No, because it is divisible by . 3. It fails the square-free condition, so it is not a Carmichael number. D) 255: 1. Prime factorization: . It is composite. 2. Is it square-free? Yes. 3. Check the divisibility condition: . * For : . Does ? No, because 254 is not divisible by 4 (). 4. Since the condition fails for , 255 is not a Carmichael number.

A linear congruence has solutions if and only if divides . If this condition is met, there are exactly incongruent solutions modulo .

In this problem, , . We are told there are exactly 3 solutions.
This implies that must be equal to 3. Let's verify this.

Prime factorization of : .
Prime factorization of : .
So, .

The problem states there are 3 solutions, but our calculation shows . This indicates a contradiction in the premise of the question. Let's adjust the question to be solvable. Let's ask for 37 solutions.

Corrected Question Premise: For which integer values of does the congruence have exactly 37 solutions modulo 185?

Since , the congruence will have exactly 37 solutions if and only if .
So, must be a multiple of 37. Now let's analyze the options based on this condition.

The question asks for the sum .
We can calculate this directly, but it's tedious. A more systematic approach is to use the provided identity or a related method.

The identity is for the sum up to . Let's calculate that first:
.
The divisors of 12 are .
Using the identity :

Summing these values: .

This sum is for from 1 to 12. The question asks for the sum from to 11.
So, we need to compute .

.

Alternatively, one could calculate each term directly:











Sum = .

The existence of a multiplicative inverse of modulo is determined by the condition .

Given that the inverse of modulo exists, we know that . This means that shares no common prime factors with . The prime factors of are .

Therefore, for each prime in the factorization of , it must be the case that does not divide . This can be written as for all .

Now consider the condition for the existence of an inverse of modulo a prime power factor . The inverse exists if and only if .
The only prime factor of is . Since we have already established that (i.e., does not divide ), it follows directly that shares no prime factors with .

Thus, for every .
This means that the multiplicative inverse of modulo must exist for each of the prime power factors of .

The other options are incorrect. The existence is not limited to , it is not independent, and it must hold for all factors, not just at least one.

This question asks for the condition of existence of solutions for a system of two linear congruences where the moduli may not be coprime. This is the generalized form of the Chinese Remainder Theorem.

The system is:

1. for some integer .
2. for some integer .

For a solution to exist, there must be an that satisfies both equations. So, we can set them equal:

Rearranging the terms, we get:

This is a linear Diophantine equation in the variables and . Such an equation has an integer solution if and only if the greatest common divisor of the coefficients of the variables divides the constant term.

The coefficients are and . Their greatest common divisor is .
The constant term is .

Therefore, a solution exists if and only if .
By the definition of modular congruence, this condition is equivalent to:
, or equivalently, .

If this condition holds, there is a unique solution modulo . The other options are either too restrictive (coprime moduli is a sufficient but not necessary condition) or incorrect.

We want to compute .
The modulus is a prime number.

By Fermat's Little Theorem, for any integer not divisible by a prime , we have . Here, . So for .

Let's consider a term in the sum for .
.
Modulo 101, this becomes:
.

This simplification applies to every term in the sum from to .
So, the sum is congruent to:

This is the sum of an arithmetic series. The sum is given by the formula .
Here, .
Sum .

Now, we need to find the remainder of this sum when divided by 101:
.

Therefore, the remainder of the sum when divided by 101 is 0.
Note that this result holds for any prime : and .

We assume the most frequent plaintext letter maps to the most frequent ciphertext letter, and likewise for the second most frequent.

• e (plaintext, x=4) maps to K (ciphertext, y=10)
• t (plaintext, x=19) maps to W (ciphertext, y=22)

This gives us a system of two linear congruences with two variables, and :

Subtract the first congruence from the second:

This congruence is of the form . A solution exists if divides . Here, , which divides 12. So a unique solution for exists.
To solve for , we need the inverse of 15 modulo 26. We can use the Extended Euclidean Algorithm or test values. Let's find .
. So, .

Multiply the congruence by 7:


. So .
. . . .
So . But $\gcd(6,26)=2
eq 1$, so this cannot be a valid key 'a' for an affine cipher.Let's re-read the problem. Maybe the mapping is swapped?Assume 'e' -> 'W' and 't' -> 'K'.

2. 
   Subtract 1 from 2: .
   Here . . . . $\gcd(20,26)=2
   eq 115a \equiv 12 \pmod{26}15a - 12 = 26k$.$3(5a - 4) = 26kkk=3j$.$3(5a-4) = 26(3j) \implies 5a-4 = 26j \implies 5a \equiv 4 \pmod{26}5 \cdot 21 = 105 = 4 \cdot 26 + 15^{-1} \equiv 21$.$a \equiv 4 \cdot 21 = 84 \pmod{26}$.$a \equiv 6 \pmod{26}a=6a=6a=3, b=24E(x) = 3x+24$.$E(4) = 3(4)+24 = 12+24=36 \equiv 10E(19) = 3(19)+24 = 57+24=81$. $81 = 3 \cdot 26 + 3 \equiv 3(3, 24)E(x) = 3x+24 \equiv 22 \pmod{26}$.$3x \equiv 22-24 = -2 \equiv 24 \pmod{26}$.$3x \equiv 24 \pmod{26}\gcd(3,26)=13^{-1}=9x \equiv 8 \cdot 9 = 72 \pmod{26}$. $72 = 2 \cdot 26 + 20$. $x=204a+b=1015a = -7 = 19$.$a = 19 \cdot 7 = 133 = 5 \cdot 26 + 3a=3$.$4(3)+b=10 \implies 12+b=10 \implies b=-2=24(3, 24)$. The question must have had a typo, where the second letter should have been 'D', not 'W'.