Easy does it! The Smart Scholar Method
The number cannot end in 1, 3 or 5, because then it would make it odd and hence not divisible by 2, let alone 8.
A number is divisible by 8 if the number formed by the rightmost three digits is divisible by 8. The number has therefore to end in 136, 152, 216, 256, 264, 312, 352, 416, 432, 456, 512, 536, 624 or 632. That is, 14 possibilities.
The first three digits from the left do not matter. For any six-digit number ending in, say, 216, we can have 3! = 3*2*1 = 6 possibilities using 3, 4 and 5.
Thus there are: 14*6 = 84 such numbers.
Siddharth is right.
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Dear Wadehra Bhai,
You have done 79ers proud. While congratulating you on this occasion let me say that by tagging ourselves (79ers)with you as a fellow 79er, we too have become "smart scholars". Smart way of snatching some credit you say?
Puzzle # 14 Easy Method > Fight with Eight
