There is a six digit code. Its first two digits, multiplied by 3 gives all ones. And the next two digits multiplied by 6 give all twos. Remaining two digits multiplied by 9 gives all threes. Then what is the code?

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pawan deep singh

  • Aug 30th, 2005

therefore the code is 373737


  • Jan 17th, 2006


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anand & giriraj

  • Feb 21st, 2006

ANSWER- 37 87 37

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prateek nayak

  • Feb 26th, 2006

1st 2 digit- 111/3=37

2nd 2 digit- 222/6=37

3rd 2 digit- 333/9=37

so the number=373737

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  • Jun 4th, 2006

It will be 373737

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R.Dharves Mohideen

  • Jul 4th, 2006

Assume the digit xx xx xx (six digits)

First Two digit       xx * 3=111


( first two digits of 1 is not divisible by 3   so we can use 111)

Second Two digit xx*6=222


( first two digits of 2 is not divisible by 6   so we can use 222)

Thrid Two digit    xx*9=333


( first two digits of 3 is not divisible by 9   so we can use 333)

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R.Dharves Mohideen

  • Jul 4th, 2006

So the Above Answer is : 373737


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  • Apr 30th, 2007

ans is 373737

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one thing is to notice is dat when first 2 digit multiply by 2 it will gv 111
so in first two digit no. ones place no. will b dat will 1 on multiplication so no. is 7
n tens digit no. will b dat will  give 11 including carry
 so first two digit will b 37

similarly other no. will b find
ans is 373737

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