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Algebra
13 Feb 2007 12:20:21 IST
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13 Feb 2007 20:46:38 IST
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read your question once again,it is not making sense
''on removing the extreme left digit of 'n', the number n/57 is also a natural number ''
when you remove first digit of n , the resultant won't be n only
so what do you mean by n/57 is a natural no. and what has it got to do with removing first digit?
, hav corrected it, srry guys
19 Feb 2007 00:39:49 IST
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Take a special case. Consider numbers 157 257 357 457 557 ....957. Removing left digits of this number will give me a number which is divisible by 57. But for all these numbers sum of digits is different. So what exactly are you looking for in the answer
25 Feb 2007 12:55:56 IST
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let the no. obtained by ommiting the digit be x then
57x=(let)abcdef????. (i)
X=bcdef??????.. (ii)
Subtracting (ii) from (i)
56x=a000000000000000??..(noif digits in no)
X=a0000000000000/56
For x to be an integer
a=7and no of digits greter than 4
x=7 x125000???.
57x=399x125000???????..
smallest such no is=49875
Depends on no of digits
Hope this is correct
25 Feb 2007 15:22:19 IST
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Let the leftmost no. be a and its palce value to be a*10b .
After removing the leftmost digit the remaining no. is divisible by 57 hence n must be odd.
As per the question : a*10b + (n/57) = n
=> 57*a*10b = 56n
=> (19).(3a).(5b).(2b-3.23) = (7).(23).n
=> (19).(3a).(5b).(2b-3) = 7n
Since n is odd then RHS will also be odd.
Hence LHS must be odd implies power of 2 should be zero.
Hence b-3 = 0 => b = 3
Putting this above we get
(19).(3a).(125) = 7n
=> 7125a = 7n
Since 7125 is not divisible by 7 and RHS has 7 as a factor then a must be 7.
Hence n comes out to be 7125.
Sum of digits = 15
Hope its clear now.
Best Wishes
After removing the leftmost digit the remaining no. is divisible by 57 hence n must be odd.
As per the question : a*10b + (n/57) = n
=> 57*a*10b = 56n
=> (19).(3a).(5b).(2b-3.23) = (7).(23).n
=> (19).(3a).(5b).(2b-3) = 7n
Since n is odd then RHS will also be odd.
Hence LHS must be odd implies power of 2 should be zero.
Hence b-3 = 0 => b = 3
Putting this above we get
(19).(3a).(125) = 7n
=> 7125a = 7n
Since 7125 is not divisible by 7 and RHS has 7 as a factor then a must be 7.
Hence n comes out to be 7125.
Sum of digits = 15
Hope its clear now.
Best Wishes
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take any multiple of 57, "add" any digit to its leftmost place