Question 26 of 31advanced🔧 ApplyShort Answer3 marks

Write a 6-digit number that it is divisible by 15, such that when the digits are reversed, it is divisible by 6.

Correct Answer

Divisibility by 15: Must end in 0 or 5, and sum of digits divisible by 3.

Divisibility by 6: Must end in even digit, and sum of digits divisible by 3.

Conditions:

  • Cannot end in 0 (reversed would start with 0, invalid for 6-digit number)
  • Must end in 5
  • Original must start with even digit
  • Sum of digits must be divisible by 3

Answer: 612345

Verification:

  • Ends in 5, sum of digits = 21 (divisible by 3)
  • ∴ 612345 is divisible by 15

Reversed: 543216

  • Ends in 6 (even), sum of digits = 21 (divisible by 3)
  • ∴ 543216 is divisible by 6
Exercise: Figure it Out (Section 4) | Q: 10 | (Page Page 22)
For More Understanding

Explanation

The chapter discusses divisibility rules. For divisibility by 15, the number must be divisible by both 3 and 5. For divisibility by 5, it must end in 0 or 5. If it ends in 0, the reversed number would start with 0, which is invalid for a 6-digit number. So it must end in 5. For the reversed number to be divisible by 6, it must be even (end in even digit) and divisible by 3. So the original number must start with an even digit. The sum of digits must be divisible by 3 for both numbers.

Solution Steps

  1. Divisible by 15: must end in 0 or 5, and sum of digits divisible by 3

  2. Cannot end in 0 (reversed would start with 0)

  3. Must end in 5

  4. Reversed must be divisible by 6: must end in even digit

  5. Original must start with even digit

  6. Sum of digits must be divisible by 3

  7. Example: 612345 → reversed 543216, both work

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