1. The prime factorization of 45 = 5 x 9 25110 is divisible by 5 as ‘0’ is at its unit place. 25110 is divisible by 9 as sum of digits is divisible by 9. Therefore, the number must be divisible by 5 x 9 = 45 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

    The prime factorization of 45 = 5 x 9
    25110 is divisible by 5 as ‘0’ is at its unit place.
    25110 is divisible by 9 as sum of digits is divisible by 9.
    Therefore, the number must be divisible by 5 x 9 = 45

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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  2. 3 + 5 = 8 and 8 is divisible by 4. 5 + 7 = 12 and 12 is divisible by 4. 7 + 9 = 16 and 16 is divisible by 4. 9 + 11 = 20 and 20 is divisible by 4. https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

    3 + 5 = 8 and 8 is divisible by 4.
    5 + 7 = 12 and 12 is divisible by 4.
    7 + 9 = 16 and 16 is divisible by 4.
    9 + 11 = 20 and 20 is divisible by 4.

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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  3. Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6. Example: (i) 2 x 3 x 4 = 24 (ii) 4 x 5 x 6 = 120 https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

    Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
    Example: (i) 2 x 3 x 4 = 24
    (ii) 4 x 5 x 6 = 120

    https://www.tiwariacademy.com/ncert-solutions/class-6/maths/chapter-3/

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    • 6