A number leaves a remainder of 1 when divided by 5 and a remainder of 2 when div
Practice Questions
Q1
A number leaves a remainder of 1 when divided by 5 and a remainder of 2 when divided by 3. What is the smallest positive integer that satisfies these conditions? (2023)
1
5
8
11
Questions & Step-by-Step Solutions
A number leaves a remainder of 1 when divided by 5 and a remainder of 2 when divided by 3. What is the smallest positive integer that satisfies these conditions? (2023)
Step 1: Understand the problem. We need to find a number that gives a remainder of 1 when divided by 5 and a remainder of 2 when divided by 3.
Step 2: Write down the conditions mathematically. Let the number be 'x'. We have two conditions: x % 5 = 1 and x % 3 = 2.
Step 3: Start testing positive integers to find the smallest number that meets both conditions. Begin with x = 1 and increase by 1 each time.
Step 4: Check x = 1: 1 % 5 = 1 (true), 1 % 3 = 1 (false). Not a solution.
Step 5: Check x = 2: 2 % 5 = 2 (false). Not a solution.
Step 6: Check x = 3: 3 % 5 = 3 (false). Not a solution.
Step 7: Check x = 4: 4 % 5 = 4 (false). Not a solution.
Step 8: Check x = 5: 5 % 5 = 0 (false). Not a solution.
Step 9: Check x = 6: 6 % 5 = 1 (true), 6 % 3 = 0 (false). Not a solution.
Step 10: Check x = 7: 7 % 5 = 2 (false). Not a solution.
Step 11: Check x = 8: 8 % 5 = 3 (false). Not a solution.
Step 12: Check x = 9: 9 % 5 = 4 (false). Not a solution.
Step 13: Check x = 10: 10 % 5 = 0 (false). Not a solution.
Step 14: Check x = 11: 11 % 5 = 1 (true), 11 % 3 = 2 (true). This is a solution.
Step 15: Since we are looking for the smallest positive integer, we continue checking lower numbers until we find the smallest one.
Step 16: Check x = 8 again: 8 % 5 = 3 (false). Not a solution.
Step 17: The smallest number that satisfies both conditions is 11.
Modular Arithmetic – Understanding how to work with remainders when dividing numbers.
System of Congruences – Solving multiple conditions that involve remainders.
