Write a 3 digit number. Repeat the same number by the side so that it becomes a 6 digit number. i.e. Supposing you wrote a number 407 then you write the same 407 again so that the resultant 6 digit number is 407407. Now you may use a calculator if you are not good at dividing.
The 6 digit number you have written is cleanly divisible by 7 without any remainder. (for example 407407 is divisble by 7 and leaves a quotient of 58201). Check it out for the your number and find the quotient. Now the resultant Quotient is divisible by 11 without any remainder. (58201 is divisble by 11 and the quotient is 5291). Check it out for your number and find the quotient again. Now the resultant quotient is divisible by 13 without any remainder. (5291 is divisible by 13 and the quotient is 407!). Check it out and are you surprised to find out the resultant quotient is the 3 digit number which originally you wrote? Nice, is n't it? Now the question is Why does this happen for any 3 digit number ? Prove it.
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u announce price money(dont give) many people try to solve the puzzle and post the answers.Puzzles are very interesting. Keep doing it and keep on doing it. I will try to complie it and put it in the form of book so that it is very helpful to younger generation. best wishes for this novel idea - vmhan
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