Question 7 of 19advanced🔧 ApplyNumerical4 marks

Find a number that leaves a remainder of 2 when divided by 3, a remainder of 3 when divided by 4, and a remainder of 4 when divided by 5. What is the smallest such number? Can you give a simple explanation of why it is the smallest?

Correct Answer

Step 1: Analyze the remainder conditions.

n≡2(mod3)→n+1≡0(mod3)n ≡ 2 (mod 3) → n + 1 ≡ 0 (mod 3) n≡3(mod4)→n+1≡0(mod4)n ≡ 3 (mod 4) → n + 1 ≡ 0 (mod 4) n≡4(mod5)→n+1≡0(mod5)n ≡ 4 (mod 5) → n + 1 ≡ 0 (mod 5)

Step 2: Determine the requirement for n+1n + 1. n+1n + 1 must be a common multiple of 3, 4, and 5.

Step 3: Calculate the Least Common Multiple (LCM). LCM(3, 4, 5) = 60

Step 4: Solve for nn. Smallest positive n+1=60n + 1 = 60

n=60−1=59n = 60 - 1 = 59

Answer: The smallest such number is 59.

Exercise: Figure it Out (Section 1) | Q: 8 | (Page Page 11)
For More Understanding

Explanation

Notice that n ≡ 2 (mod 3) means n + 1 ≡ 0 (mod 3). Similarly, n ≡ 3 (mod 4) means n + 1 ≡ 0 (mod 4), and n ≡ 4 (mod 5) means n + 1 ≡ 0 (mod 5). So n + 1 must be a common multiple of 3, 4, and 5. The LCM of 3, 4, and 5 is 60. Therefore, n + 1 = 60, giving n = 59. This is the smallest positive solution because LCM(3,4,5) = 60 is the smallest positive common multiple.

Solution Steps

  1. Step 1: n ≡ 2 (mod 3) → n + 1 ≡ 0 (mod 3)

  2. Step 2: n ≡ 3 (mod 4) → n + 1 ≡ 0 (mod 4)

  3. Step 3: n ≡ 4 (mod 5) → n + 1 ≡ 0 (mod 5)

  4. Step 4: n + 1 must be a common multiple of 3, 4, and 5

  5. Step 5: LCM(3, 4, 5) = 60

  6. Step 6: Smallest positive n + 1 = 60, so n = 59

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