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13 Σεπ 2015 · Practice Problem Solutions 1. Given that 5x 6 (mod 8), nd x. [Solution: 6] 2. Find the last digit of 7100 [Solution: 1] 7100 (72) 50 49 ( 1)50 1 mod 10. 3. (1992 AHSME 17) The two-digit integers form 19 to 92 are written consecutively to form the large integer N = 192021 909192. Suppose that 3k is the highest power of 3 that is a factor of N ...
Ex 6: We can solve the equation 3 · x + 6 ≡ 8(mod 10) by using the sum (3) and multiplication (4) rules along with the above table: 3·x+6 ≡ 8(mod 10) =⇒
Basic Practice. Compute the modular arithmetic quantities, modulo n, in such a way that your answer is an integer 0 k < n. Do NOT use a calculator. Do these in your head. Compare to the answer key at the end.
24 Μαΐ 2024 · Using a Calculator. For calculating 1258 in modulo 4 (as ‘A’ in modulo ‘n’) using a standard calculator, we follow the following steps: Step 1: Dividing ‘A’ by ‘n’ 1258 ÷ 4 = 314.5. Step 2: Subtracting the whole Part of the Resulting Quantity. 314.5 – 314 = 0.5. Step 3: Multiplying the Difference by ‘n’ to Obtain the Modulus
In addition to clock analogy, one can view modular arithmetic as arithmetic of remain-ders. For example, in mod 12 arithmetic, all the multiples of 12 (i.e., all the numbers that give remainder 0 when divided by 12) are equivalent to 0. In the modular arithmetic notation, this can be written as.
Proof If x3+ 10000 = y3then x3+ 10000 ≡ y3(mod 7) (by Theorem 2.1.3(1)). Since 10000 ≡ 4 (mod 7), x3+4 ≡ y3(mod 7). But x3≡ −1,0, or 1 (mod 7) by previous example, so x3+ 4 ≡ 3,4 or 5 (mod 7), while y3≡ −1,0, or 1 (mod 7) contradiction. This example illustrates one of the uses of modular arithmetic.
Problem 1: just an example of previous identity. Find the remainder of 46 23 (a.k.a. 46 .23) on division by 7. Solution. 46 4 (mod 7) and 23 2 (mod 7) so 46 .23 4 .2 8 1 (mod 7). Problem 2: look for a small partner. Compute 138 (mod 7). What does this mean? The unique remainder r satisfying 0 r < 7. Easiest solution. 138 ( 1)8 1 (mod 7).
