If a number leaves a remainder of 2 when divided by 4 and a remainder of 3 when

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Question: If a number leaves a remainder of 2 when divided by 4 and a remainder of 3 when divided by 5, what is the smallest such number?

Options:

Correct Answer: 7

Solution:

The smallest number that satisfies both conditions is 7 (7 % 4 = 3 and 7 % 5 = 2).

If a number leaves a remainder of 2 when divided by 4 and a remainder of 3 when divided by 5, what is the smallest such number?

If a number leaves a remainder of 2 when divided by 4 and a remainder of 3 when divided by 5, what is the smallest such number?

Step 1: Understand the problem. We need to find a number that gives a remainder of 2 when divided by 4 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 4, we can express this as: x % 4 = 2.
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 % 5 = 3.
Step 4: List some numbers that satisfy the first condition (x % 4 = 2). These numbers are: 2, 6, 10, 14, 18, ... (add 4 each time).
Step 5: Check each of these numbers to see if they also satisfy the second condition (x % 5 = 3).
Step 6: Start with the first number from the list: 2. Check: 2 % 5 = 2 (not a match).
Step 7: Move to the next number: 6. Check: 6 % 5 = 1 (not a match).
Step 8: Move to the next number: 10. Check: 10 % 5 = 0 (not a match).
Step 9: Move to the next number: 14. Check: 14 % 5 = 4 (not a match).
Step 10: Move to the next number: 18. Check: 18 % 5 = 3 (this is a match!).
Step 11: Verify that 18 also satisfies the first condition: 18 % 4 = 2 (this is also a match!).
Step 12: Conclude that the smallest number that satisfies both conditions is 18.

Modular Arithmetic – Understanding how to work with remainders when dividing numbers.
System of Congruences – Solving multiple conditions that involve remainders.