Hi,
Recently I came across a good puzzle. It says "what is lowest possible number which when divided by 2,3,5,7,9 and 11 leaves remainder 1,2,3,4,5 and 6 respectively".
In this puzzle method and approach is more important than answer.
Give it a shot.
Agry
The answer can easily derived by trial and error method but the important part is logic.
ReplyDelete11 is the highest divisor so we will start with it and we will add 6(remainder) to the answers.
1) We will see at what point do 11 and 9 meet (We have taken highest numbers to reduce the calculation steps)
11 * 1 = 11 now 11 + 6 (Remainder) = 17 and 17 - 5 = 12 which is not divisible by 9
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11* 4 = 44 => 44 + 6 = 50 => 50-5 = 45 which is divisible by 9
Now 50 will be the common added and 9 * 11 will be the added there after
2) We will see at what point 7 will also meet 11 & 9
50 + 99(i.e. 11*9) = 149 => 149 - 4/7 = 20.71
149 + 99 = 248 => 248 -4/7 = 34.85
248 + 99 = 347 => 347 - 4/7 = 49
Therefore 11, 9 & 7 meet at 347
3) We will see at what point 5 will meet 11, 9 & 7
347 + 693 (11*9*7) = 1040 => 1040 - 3/5 = 207.4
1040 + 693 = 1733 => 1733 -3/5 = 346
Therefore 11, 9, 7 & 5 meet at 1733
4) As 1733 meets 9 then it automatically meets 3 too so no need to check
5) 1733 -1/2 = 866
Hence the Minimum number at which all the numbers will meet the condition is 1733.