IIT JEE , AIEEE Forum - Mathematics , Algebra

magiclko

find the sum of all the digits of a natural number n such that on removing the extreme left digit of 'n', the number n/57 is also a natural number

malay

157,114 etc.
take any multiple of 57, "add" any digit to its leftmost place

AAKRITI

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?

mukul1yellow9

hey.manasi idon't understand ques. plz make it clear to me

magiclko

wat i mean to say is, suppose n is 234578, then on removing 2, 34578 shud be divisible by 57

note: the question was a bit wrong

, hav corrected it, srry guys

krishna.gopal

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

magiclko

the question given was this much only, nothin else.... it was in some coaching entrance test.... may be they would be asking some general form !!!!

rik_mad

the number is 7125 and sum is 15

pink_ele

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

pink_ele

c'mon what do you say

rik_mad

hey i hav already posted the smallest number it is ..7125
see remove 7 u get 125 which is nothing but 7125/57
rite...!!!

iitkgp_bipin

Let the leftmost no. be a and its palce value to be a*10^b .

After removing the leftmost digit the remaining no. is divisible by 57 hence n must be odd.

As per the question :   a*10^b + (n/57) = n

=>   57*a*10^b = 56n

=>   (19).(3a).(5^b).(2^b-3.2³) = (7).(2³).n

=>   (19).(3a).(5^b).(2^b-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

umang

oh !

I wrongly understood the question !!!

yes , your answer is correct !!!!!

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