Question
Find the difference between minimum and maximum value
of 'g' such that '9g5621' is always divisible by 3.Solution
A number is divisible by 3 when the sum of its digits is divisible by 3.
Sum of digits of '9g5621' = (9 + g + 5 + 6 + 2 + 1) = g + 23.
So, g + 23 should be divisible by 3.
Possible values of 'g' = 2, 5, 8
Minimum value of 'g' = 2
Maximum value of 'g' = 8
Required difference = 8 – 2 = 6
64, 65, 73, 100, ?, 289
What will come in place of the question mark (?) in the following series?
325, 292, 262, 229, ?, 166
?, 6, 6, 9, 18, 45
16, 15, 24, -1, 48, ?
16, 32, 68, 132, ?, 376
452, 369, ?, 217, 146, 79
12    48      192      ?      3072       12288
...What will come in place of the question mark (?) in the following series?
100, 729, 64, 343, 36, ?
What will come in place of the question mark (?) in the following series?
10, 135, ?, 226, 234, 235
79, 248, ?, 225, 125, 206
