📢 Too many exams? Don’t know which one suits you best? Book Your Free Expert 👉 call Now!


    Question

    An eight-digit number, 6x2y824 is divisible by 72. Find

    the maximum possible value of x + y.
    A 15 Correct Answer Incorrect Answer
    B 14 Correct Answer Incorrect Answer
    C 13 Correct Answer Incorrect Answer
    D 12 Correct Answer Incorrect Answer

    Solution

    Given number is 6x2y824. For a number to be divisible by 72, then it should be divisible by both 8 and 9. So, we will check the divisibility rule of 8 and 9. For the given number to be divisible by 8, the last three digits of the number should be divisible by 8. Last three digits of the given number are 824. So, the possible values of y are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (since 824 is already divisible by 8). For the number to be divisible by 9, the sum digits of the given number should be a multiple of 9. So, sum of the digits of the given number
    = 6 + x + 2 + y + 8 + 2 + 4 = 22 + x + y So, next multiples of 9 after 22 are 27 and 36. So, possible values of (x + y) are 5 and 14. When, x + y = 14, take y = 9 and x = 5. So, this case is valid and sufficient to conclude that maximum value of x + y = 14. Hence, option B.

    Practice Next
    ask-question