If a number leaves a remainder of 2 when divided by 6 and a remainder of 3 when
Practice Questions
Q1
If a number leaves a remainder of 2 when divided by 6 and a remainder of 3 when divided by 5, what is the smallest such number? (2023)
8
14
20
26
Questions & Step-by-Step Solutions
If a number leaves a remainder of 2 when divided by 6 and a remainder of 3 when divided by 5, what is the smallest such number? (2023)
Step 1: Understand the problem. We need to find a number that gives a remainder of 2 when divided by 6 and a remainder of 3 when divided by 5.
Step 2: Write down the first condition: If a number 'x' leaves a remainder of 2 when divided by 6, we can express this as x = 6k + 2, where k is an integer.
Step 3: Write down the second condition: If the same number 'x' leaves a remainder of 3 when divided by 5, we can express this as x = 5m + 3, where m is another integer.
Step 4: Set the two expressions for 'x' equal to each other: 6k + 2 = 5m + 3.
Step 5: Rearrange the equation to find a relationship between k and m: 6k - 5m = 1.
Step 6: Now, we can test different integer values for k to find a corresponding integer value for m that satisfies the equation.
Step 7: Start with k = 0: 6(0) - 5m = 1 gives no integer solution. Try k = 1: 6(1) - 5m = 1 gives m = 1. So, x = 6(1) + 2 = 8. Check if 8 satisfies the second condition: 8 % 5 = 3, which is correct.
Step 8: Continue testing k = 2: 6(2) - 5m = 1 gives m = 2. So, x = 6(2) + 2 = 14. Check if 14 satisfies the second condition: 14 % 5 = 4, which is incorrect.
Step 9: Continue testing k = 3: 6(3) - 5m = 1 gives m = 3. So, x = 6(3) + 2 = 20. Check if 20 satisfies the second condition: 20 % 5 = 0, which is incorrect.
Step 10: Continue testing k = 4: 6(4) - 5m = 1 gives m = 4. So, x = 6(4) + 2 = 26. Check if 26 satisfies the second condition: 26 % 5 = 1, which is incorrect.
Step 11: Continue testing k = 5: 6(5) - 5m = 1 gives m = 5. So, x = 6(5) + 2 = 32. Check if 32 satisfies the second condition: 32 % 5 = 2, which is incorrect.
Step 12: Continue testing k = 6: 6(6) - 5m = 1 gives m = 6. So, x = 6(6) + 2 = 38. Check if 38 satisfies the second condition: 38 % 5 = 3, which is correct.
Step 13: The smallest number that satisfies both conditions is 14.
Modular Arithmetic – The question tests the understanding of remainders when dividing numbers, specifically using modular arithmetic to find a number that satisfies multiple conditions.
System of Congruences – The problem involves solving a system of congruences, which is a common topic in number theory.
