Question
If ‘1x5629’ is a six digit number which is divisible
by 9, then which of the following can be the minimum value of ‘x’?Solution
For a number to be divisible by ‘9’, the sum of its digits must be divisible by ‘9’.So, (1 + x + 5 + 6 + 2 + 9) = (23 + x)So, number after 23 which is multiple of 9 is 27,Therefore, minimum value of ‘x’ = 27 – 23 = 4
? = (597.98 ÷ (6.97 2.01 – 3.1)) × 12.9
1153, 1206, ?, 1326, 1393, 1464
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