Question
Download Solution PDFIf an 8-digit number 1a9759b0 is divisible by 108, then the maximum value of (7a + 3b) is:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
The 8-digit number is 1a9759b0.
The number is divisible by 108.
108 = 4 × 27
Formula Used:
Divisibility rule of 4: The number formed by the last two digits must be divisible by 4.
Calculation:
For divisibility by 4, the last two digits 'b0' must be divisible by 4. Possible values of b are 0, 2, 4, 6, 8.
Taking b = 8 for maximum value of (7a + 3b).
Now the number is: 1a975980
Sum of each digit = 1 + a + 9 + 7 + 5 + 9 + 8 + 0 = 39 + a
Since a is since digit number, 39 + a must be divisible by 9.
39 + a = 45
a = 6
Maximum value of (7a + 3b) = (6 × 7 + 8 × 3) = 66
The maximum value of (7a + 3b) is 66.
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