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QUESTIONS 1. You need to refer to the sections on Congruence and Arithmetic mod...
Jun 5, 2024
Solution by Steps
step 1
To show that 10≡3110 \equiv_{3} 1, we start by noting that 10=3×3+110 = 3 \times 3 + 1. Therefore, 10≡1(mod3)10 \equiv 1 \pmod{3}
step 2
For any positive integer kk, 10k≡31k=110^k \equiv_{3} 1^k = 1. This follows from the property of congruences that if a≡b(modm)a \equiv b \pmod{m}, then ak≡bk(modm)a^k \equiv b^k \pmod{m}
step 3
To show that A×10k≡3AA \times 10^k \equiv_{3} A, we use the result from step 2. Since 10k≡1(mod3)10^k \equiv 1 \pmod{3}, we have A×10k≡A×1=A(mod3)A \times 10^k \equiv A \times 1 = A \pmod{3}
step 4
Consider a number NN with digits [ak,ak−1,⋯ ,a1,a0][a_k, a_{k-1}, \cdots, a_1, a_0]. This number can be written as N=ak×10k+ak−1×10k−1+⋯+a1×10+a0N = a_k \times 10^k + a_{k-1} \times 10^{k-1} + \cdots + a_1 \times 10 + a_0
step 5
Using the result from step 3, we have ai×10i≡ai(mod3)a_i \times 10^i \equiv a_i \pmod{3} for each digit aia_i. Therefore, N≡ak+ak−1+⋯+a1+a0(mod3)N \equiv a_k + a_{k-1} + \cdots + a_1 + a_0 \pmod{3}
step 6
Since N≡ak+ak−1+⋯+a1+a0(mod3)N \equiv a_k + a_{k-1} + \cdots + a_1 + a_0 \pmod{3}, NN is divisible by 3 if and only if the sum of its digits ak+ak−1+⋯+a1+a0a_k + a_{k-1} + \cdots + a_1 + a_0 is divisible by 3
step 7
A number is divisible by 6 if and only if it is divisible by both 2 and 3. Therefore, a test for divisibility by 6 is to check if the number is even (last digit is 0, 2, 4, 6, or 8) and if the sum of its digits is divisible by 3
Answer
The test for divisibility by 3 works because 10k≡1(mod3)10^k \equiv 1 \pmod{3}, and a number is divisible by 3 if the sum of its digits is divisible by 3. A test for divisibility by 6 is to check if the number is even and if the sum of its digits is divisible by 3.
Key Concept
Divisibility rules
Explanation
The divisibility rule for 3 is based on the property that 10k≡1(mod3)10^k \equiv 1 \pmod{3}. For 6, a number must be divisible by both 2 and 3.
Solution by Steps
step 1
To show that 10≡3110 \equiv_{3} 1, we start by noting that 10=3×3+110 = 3 \times 3 + 1. Therefore, 10≡1(mod3)10 \equiv 1 \pmod{3}
step 2
For any positive integer kk, 10k≡31k=110^k \equiv_{3} 1^k = 1. This follows from the property of congruences that if a≡b(modn)a \equiv b \pmod{n}, then ak≡bk(modn)a^k \equiv b^k \pmod{n}
step 3
To show that A×10k≡3AA \times 10^k \equiv_{3} A for any positive integers kk and AA, we use the result from step 2. Since 10k≡1(mod3)10^k \equiv 1 \pmod{3}, we have A×10k≡A×1=A(mod3)A \times 10^k \equiv A \times 1 = A \pmod{3}
step 4
Consider a number NN with digits [ak,ak−1,⋯ ,a1,a0][a_k, a_{k-1}, \cdots, a_1, a_0]. This number can be written as N=ak×10k+ak−1×10k−1+⋯+a1×10+a0N = a_k \times 10^k + a_{k-1} \times 10^{k-1} + \cdots + a_1 \times 10 + a_0
step 5
Using the results from steps 2 and 3, we have N≡3ak+ak−1+⋯+a1+a0N \equiv_{3} a_k + a_{k-1} + \cdots + a_1 + a_0. Therefore, NN is divisible by 3 if and only if the sum of its digits ak+ak−1+⋯+a1+a0a_k + a_{k-1} + \cdots + a_1 + a_0 is divisible by 3
step 6
For divisibility by 6, a number must be divisible by both 2 and 3. A number is divisible by 2 if its last digit is even. Therefore, a number is divisible by 6 if the sum of its digits is divisible by 3 and its last digit is even
Answer
A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 6 if the sum of its digits is divisible by 3 and its last digit is even.
Key Concept
Divisibility rules for 3 and 6
Explanation
The key concept is that a number's divisibility by 3 can be determined by the sum of its digits, and for divisibility by 6, the number must also have an even last digit.
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