IIT JEE , AIEEE Forum - Mathematics , Algebra

freestyler_fx

let abcdef be a number....

if we multiply it with 6 it becomes defabc..

find the numberz

rik_mad

hi,

abcdef can be written as (abc)10³ + def

and we can write defabc in the same manner

by condition

6[(abc)10³ + def] = (def)10³+ abc

5999(abc) = (def)994

simplifies to

857(abc) = 142(def)

now u cannot simplify the numbers further (i.e. 142 and 857)

so clearly 'abc' must be a multiple of '142' and def of 857

taking 'abc' as 142 and 'def' as 857 we get number 142857

and on verifying we find 142857*6 = 857142

so number is 142857

iitkgp_bipin

rik_mad has done a great job.

If we multiply any number by 6 then at most its last three digits are altered.

Hence 6[(abc)10³ + def] = [(def)10³+ abc]

=> 857(abc) = 142(def)

Hence abc = 142 and def = 857

Hence the orignal no. = 142857

anit_sahu

i would like to mention something about this number.this number is known as revolving numbers.

142857*2=285714

142857*3=428571

142857*4=571428

142857*5=714285

142857*6=857142

142857*7=999999

1/7=0.142857142857.......

2/7=0.428571428571.......

3/7=0.571428571428.......

and so on.

this identical question was also there in fiitjee open test fot iit jee 2008

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