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
148, ‘?’ , 15 4 , 17 8 , 23 8 , 35 8 , 56 8
3601 3602 1803 604 154 36
...8, 7, 12, 35, 128, 635
14, 28, 54, 98, 154, 224
104Â Â Â Â Â 136Â Â Â Â Â 152Â Â Â Â 160Â Â Â Â Â 164Â Â Â Â Â ?
...A’ is the nth term of the given series and ‘B’ is the (n + 1)th term of the given
series. Choose the correct statement from the following s...
7    7       14    42   ?     840
...11Â Â Â 12.1Â Â Â Â 14.3Â Â Â Â 17.6Â Â Â Â Â 22Â Â Â Â Â ?
9Â Â Â 4.5Â Â Â 4.5Â Â Â 9Â Â Â 36Â Â Â ?
40 30 20 ? 7.5 4.375
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