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Problem 6: If
a-=b(mod m) and
b-=c(mod m), show that
a-=c(mod m).
Problem 7: Use the Fundamental Theorem of Arithmetic to give a different proof of Euclid's Lemma. (Hint: you will need to use both the existence and uniqueness of prime number factorization.)
Problem 8: Suppose
a-=b(mod m). Show that
c is a common divisor of
a and
m if and only if
c is a common divisor of
b and
m. (This in particular implies that the greatest common divisor
(a,m)=(b,m).
Problem 9: Show that a natural number is divisible by 3 if and only if the sum of all its digits (in base 10) is divisible by 3 .

Problem $6$ : If $a \equiv b(\bmod m)$ and $b \equiv c(\bmod m)$ , show that $a \equiv c(\bmod m)$ . $\newline$ Problem $7$ : Use the Fundamental Theorem of Arithmetic to give a different proof of Euclid's Lemma. (Hint: you will need to use both the existence and uniqueness of prime number factorization.) $\newline$ Problem $8$ : Suppose $a \equiv b(\bmod m)$ . Show that $c$ is a common divisor of $a$ and $m$ if and only if $c$ is a common divisor of $b$ and $m$ . (This in particular implies that the greatest common divisor $b \equiv c(\bmod m)$ $0$ . $\newline$ Problem $9$ : Show that a natural number is divisible by $3$ if and only if the sum of all its digits (in base $10$ ) is divisible by $3$ .

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Math Problems

Grade 7

One-step inequalities: word problems

Full solution

Q. Problem

6

: If

a \equiv b(\bmod m)

and

b \equiv c(\bmod m)

, show that

a \equiv c(\bmod m)

\newline

Problem

7

: Use the Fundamental Theorem of Arithmetic to give a different proof of Euclid's Lemma. (Hint: you will need to use both the existence and uniqueness of prime number factorization.)

\newline

Problem

8

: Suppose

a \equiv b(\bmod m)

. Show that

c

is a common divisor of

a

and

m

if and only if

c

is a common divisor of

b

and

m

. (This in particular implies that the greatest common divisor

b \equiv c(\bmod m)

0

\newline

Problem

9

: Show that a natural number is divisible by

3

if and only if the sum of all its digits (in base

10

) is divisible by

3

Use Modular Arithmetic: Since $a\equiv b\pmod{m}$ , we can write $a = b + km$ for some integer $k$ .
Apply Euclid's Lemma: Similarly, since $b \equiv c \pmod{m}$ , we can write $b = c + lm$ for some integer $l$ .
Utilize Fundamental Theorem: Substitute the second equation into the first to get $a = (c + lm) + km$ .
Prove Common Divisors: Combine like terms to get $a = c + (k + l)m$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a ext{−} c ext{ (mod } m)$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ .The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers.
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else.
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \mod m$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \mod m$ $0$ for some integer $a \equiv c \mod m$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \mod m$ $4$ and $a \equiv c \mod m$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \mod m$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \mod m$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \mod m$ $0$ for some integer $a \equiv c \mod m$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \mod m$ $4$ and $a \equiv c \mod m$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \mod m$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \mod m$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \mod m$ $0$ for some integer $a \equiv c \mod m$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \mod m$ $4$ and $a \equiv c \mod m$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \mod m$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \mod m$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \mod m$ $0$ for some integer $a \equiv c \mod m$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \mod m$ $4$ and $a \equiv c \mod m$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \mod m$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ . Each term except the last is divisible by $b$ $2$ because $b$ $3$ is congruent to $1$ modulo $b$ $2$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ . Each term except the last is divisible by $b$ $2$ because $b$ $3$ is congruent to $1$ modulo $b$ $2$ . The sum of the digits is $b$ $6$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \mod m$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \mod m$ $0$ for some integer $a \equiv c \mod m$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \mod m$ $4$ and $a \equiv c \mod m$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \mod m$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ . Each term except the last is divisible by $b$ $2$ because $b$ $3$ is congruent to $1$ modulo $b$ $2$ . The sum of the digits is $b$ $6$ . If $a$ $8$ is divisible by $b$ $2$ , then the sum of the digits must also be divisible by $b$ $2$ , since the non-unit terms are already multiples of $b$ $2$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ . Each term except the last is divisible by $b$ $2$ because $b$ $3$ is congruent to $1$ modulo $b$ $2$ . The sum of the digits is $b$ $6$ . If $a$ $8$ is divisible by $b$ $2$ , then the sum of the digits must also be divisible by $b$ $2$ , since the non-unit terms are already multiples of $b$ $2$ . Conversely, if the sum of the digits is divisible by $b$ $2$ , then $a$ $8$ must be divisible by $b$ $2$ because the difference between $a$ $8$ and the sum of its digits is a multiple of $b$ $2$ .
Examine Digit Divisibility: Since $k + l$ is also an integer, we have shown that $a \equiv c \pmod{m}$ . The Fundamental Theorem of Arithmetic states that every integer greater than $1$ can be written uniquely as a product of prime numbers. Euclid's Lemma states that if a prime $p$ divides the product $ab$ , then $p$ must divide $a$ or $b$ . Let's assume $p$ divides $ab$ , which means $a \equiv c \pmod{m}$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . By the Fundamental Theorem, we can express $a$ and $b$ as unique products of primes: $a \equiv c \pmod{m}$ $4$ and $a \equiv c \pmod{m}$ $5$ . If $p$ is not a factor of $a$ , then it must be a factor of $b$ , because the prime factorization of $ab$ must include $p$ , and the uniqueness part of the theorem ensures that $p$ can't come from anywhere else. Let $1$ $2$ be a common divisor of $a$ and $1$ $4$ , so $1$ $2$ divides $a$ and $1$ $2$ divides $1$ $4$ . Since $1$ $9$ , we have $p$ $0$ for some integer $a \equiv c \pmod{m}$ $1$ . If $1$ $2$ divides $a$ , then $p$ $4$ for some integer $p$ $5$ . Substitute $a$ into the congruence relation to get $p$ $7$ . Rearrange to $p$ $8$ . Since $1$ $2$ divides both terms on the right side, $1$ $2$ must divide $b$ . Therefore, $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Now, let's prove the converse: if $1$ $2$ is a common divisor of $b$ and $1$ $4$ , then $1$ $2$ is a common divisor of $a$ and $1$ $4$ . If $1$ $2$ divides $b$ , then $p$ $3$ for some integer $p$ . Substitute $b$ into the congruence relation to get $p$ $6$ . Since $1$ $2$ divides $1$ $4$ , it also divides $p$ $9$ . Therefore, $1$ $2$ divides $a$ . We have shown that $1$ $2$ is a common divisor of $a$ and $1$ $4$ if and only if $1$ $2$ is a common divisor of $b$ and $1$ $4$ . Let the natural number be $a$ $8$ , and let its digits be $a$ $9$ . $a$ $8$ can be written as $b$ $1$ . Each term except the last is divisible by $b$ $2$ because $b$ $3$ is congruent to $1$ modulo $b$ $2$ . The sum of the digits is $b$ $6$ . If $a$ $8$ is divisible by $b$ $2$ , then the sum of the digits must also be divisible by $b$ $2$ , since the non-unit terms are already multiples of $b$ $2$ . Conversely, if the sum of the digits is divisible by $b$ $2$ , then $a$ $8$ must be divisible by $b$ $2$ because the difference between $a$ $8$ and the sum of its digits is a multiple of $b$ $2$ . Therefore, $a$ $8$ is divisible by $b$ $2$ if and only if the sum of its digits is divisible by $b$ $2$ .