Question
Find the smallest 4-digit number which leaves a
remainder 5 when divided by each of 6, 8 and 9.Solution
If N leaves remainder 5 when divided by 6, 8 and 9, then N − 5 is divisible by all three. LCM(6, 8, 9): 6 = 2 × 3 8 = 2³ 9 = 3² LCM = 2³ × 3² = 8 × 9 = 72 So N − 5 = 72k N = 72k + 5 We want smallest 4-digit N ≥ 1000: 72k + 5 ≥ 1000 ⇒ 72k ≥ 995 ⇒ k ≥ 995/72 ≈ 13.8 So k = 14 N = 72 × 14 + 5 = 1,008 + 5 = 1,013 Answer: 1,013.
226  231   466  ?  5606  28031    Â
What will come in place of the question mark (?) in the following series?
80, 200, 400, 600, ?, 300
What will come in place of the question mark (?) in the following series?
72, 108, 172, ?, 416, 612
What will come in place of the question mark (?) in the following series?
175, 160, 140, 115, ?
44   45    41   50    ?      59    23
...24    12    ?    24  96    768
168, 164, 155, 139, ?, 78
380, ?, 360, 324, 260, 160
196, 196, 203, 229, 292, ?
79, 248, ?, 225, 125, 206
Relevant for Exams:
