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Algebra
23 Feb 2007 16:52:54 IST
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pink_ele
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23 Feb 2007 22:46:12 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....(no.of digits in number)
X=a0000000000000/56
For x to be an integer
a=7and no of digits greter than 4
x=7000....
57x=399000.........
Depends on no of digits
Hope this is correct
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23 Feb 2007 22:47:49 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=7000???.
57x=399000???????..
Depends on no of digits
Hope this is correct
...even i hav asked it once
24 Feb 2007 16:57:24 IST
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Hi pink_ele
You have nicely put ur brain but with the results i found some discrepancy as u obtained
x=7000???.
57x=399000???????..
now read the question it says
a natural number n is such that if the left most digit of the number is removed its
value becomes n/57, then what is the sum of digits of this number???
value becomes n/57, then what is the sum of digits of this number???
Now if u drop the leftmost digit from the number which is 57x = 399000...???
it becomes 990000......... which is not equal to x = 7000????
anyways good attempt keep it up
at the same time to nishantsingh89 i am trying out its solution.
24 Feb 2007 17:06:09 IST
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may be i have not understood your question properly.because i am a bit doubtfull about my interpretation of the statement "its value becomes n/57"
but if you take each of the numbers given and apply to it the condition ,none of the options seem to satisfy it
for example take n=15 .remove the leftmost digit i.e 1
then its valur becomes 5 which is not equal to n/57 i.e 15/57
same hold true for other options as well.
so plzz. reply if i am making a mistake anywhere
24 Feb 2007 22:15:40 IST
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let number be
a(10^n) + b(10^n-1) ....
now according to given condition
[a(10^n) + b(10^n-1) ....]/57 = b(10^n-1) +....
this implies
a(10^n) = 56[b(10^n-1) +....] (1)
this simply means that 56 on multiplying to a certain number
gives a0000000...
let the number in square brackets in RHS of (1)
be xyzwrt
so 56 * xyzwrt = xyzwrt
* 56
now there is only one value of t which gives 0 with six in multiplication
i'e 5 since 5*6 = 30 (3 goes in carry over)
thinking in a similar way we can asses that if r is 2 then we will get a 0 again
one can continue like this to get the complete number
but the options clearly suggest the number isn't very big so i just started putting arbitrary numbers from 0 to 9 before 25 to get that 125*56 = 7000
that did it !
i don't know if their is any other method to solve this
25 Feb 2007 12:51:39 IST
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sorry i commited a mistake
here is the ans
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=1250000???.
57x=7125000???????..
smallest such no. is57x=7125
Depends on no of digits
Hope this is correct
25 Feb 2007 15:08:57 IST
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HI
I FIGURED OUT THE SOLUTION MYSELF WHICH I AM POSTING HERE
LET,
N= a(10^x-1)+ b*(10^x-2)...............................................................---- (1)
N/57= b*(10^x-2)............................................................... ----(2)
(2) - (1)
56N/57= a(10^x-1)
basically 56n/57= a000000000000000000..............
therefore,
n= 57* (a00000000000000000000000000.......) / 56
or
n= 57*(a00000000000000000000000000...)/7*8
since n is a natural number therefore a00000000000... must be divisible by 56, i.e 7*8 (57= 19*3 it doesnt have any common factors as that of 56)
therefore a must equal to 7
n= 57*(700000000000000..........)/7*8
now to be divisible by 8 and be an integral power of 10 , such smallest number is 1000
n= 57*(7* 10000000000000000...)/7*8
cancelling the common terms
n= 57* 12500000000000000....
n= 7125000000000000000....
now we have to calculate sum of digits which will be
7+1+2+5+0+0+0+..........=15
thus 15 is the answer ,
thanks for help
now
I FIGURED OUT THE SOLUTION MYSELF WHICH I AM POSTING HERE
LET,
N= a(10^x-1)+ b*(10^x-2)...............................................................---- (1)
N/57= b*(10^x-2)............................................................... ----(2)
(2) - (1)
56N/57= a(10^x-1)
basically 56n/57= a000000000000000000..............
therefore,
n= 57* (a00000000000000000000000000.......) / 56
or
n= 57*(a00000000000000000000000000...)/7*8
since n is a natural number therefore a00000000000... must be divisible by 56, i.e 7*8 (57= 19*3 it doesnt have any common factors as that of 56)
therefore a must equal to 7
n= 57*(700000000000000..........)/7*8
now to be divisible by 8 and be an integral power of 10 , such smallest number is 1000
n= 57*(7* 10000000000000000...)/7*8
cancelling the common terms
n= 57* 12500000000000000....
n= 7125000000000000000....
now we have to calculate sum of digits which will be
7+1+2+5+0+0+0+..........=15
thus 15 is the answer ,
thanks for help
now
25 Feb 2007 15:18:02 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
25 Feb 2007 21:58:42 IST
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sorry, my method is correct and easier but
i commited a mistake, as its my habit
here is the ans
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=1250000...
57x=7125000....
smallest such no. is57x=7125
cheers!
i am sorry again
Depends on no of digits
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