Question 15 of 18advanced🔧 ApplyNumerical3 marks

Find the smallest number that is divisible by 3, 4, 5 and 7, but leaves a remainder of 10 when divided by 11.

Correct Answer

Step 1: Find LCM(3, 4, 5, 7) LCM(3, 4, 5, 7) = 3×43 \times 4 × 5×75 \times 7 = 420 (since they are co-prime)

Step 2: Check if 420 satisfies the remainder condition 420 ÷ 11 = 38 remainder 2 We need remainder 10, so we need to add 8 (since 10 − 2 = 8) 420 + 8 = 428 But 428 ÷ 5 = 85.6 (not divisible by 5)

Step 3: Check multiples of 420 modulo 11 Multiples of 420: 420, 840, 1260, 1680, 2100... 420 mod 11 = 2 840 mod 11 = 4 1260 mod 11 = 6 1680 mod 11 = 8 2100 mod 11 = 10

Answer: The smallest number is 2100

Exercise: Figure it Out (Section 6) | Q: 17 | (Page Page 18)
For More Understanding

Explanation

The number must be a multiple of LCM(3, 4, 5, 7) = 420 and leave remainder 10 when divided by 11. We check multiples of 420: 420 mod 11 = 2, 840 mod 11 = 4, 1260 mod 11 = 6, 1680 mod 11 = 8, 2100 mod 11 = 10. The smallest such number is 2100.

Solution Steps

  1. Step 1: Find LCM(3, 4, 5, 7) = 420.

  2. Step 2: Check multiples of 420 modulo 11.

  3. Step 3: 420 mod 11 = 2, 840 mod 11 = 4, 1260 mod 11 = 6, 1680 mod 11 = 8, 2100 mod 11 = 10.

  4. Step 4: The smallest number is 2100.

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