Answer
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Hint: For, this question we will question statement as $ 6\cdot (abcdef)=defabc $ and then we will just separate the digits on both side in abc and def digits and by collecting similar digits on same side we will compare the number by comparing their decimal places and hence we will add the terms.
Complete step-by-step answer:
In this question it is given that there is a six digit number abcdef and when this number is multiplied with 6 then the resulting number is also a six digit number and is defabc. Basically, we have to find the sum of all six numbers that is a + b + c + d + e +f
So, we can denote above statement as
$ 6\cdot (abcdef)=defabc $
As a, b, c, d, e, f are numbers so we can write abcdef as $ 1000\cdot abc+def $ and defabc as $ 1000.def+abc $.
$ 6\cdot (1000\cdot abc+def)=1000.def+abc $
On solving by opening brackets we get
$ 6000\cdot abc+6def=1000.def+abc $
Taking terms of abc on left side and terms of def on right side, we get
$ 6000\cdot abc-abc=1000.def-6def $
On solving, we gte
$ 5999\cdot abc=994\cdot def $
On simplifying coefficients of terms abc and def, we gte
$ 857\cdot abc=142\cdot def $
Now, just see the pattern, what we see is on left side the, in 857abc, 857 denotes def and on right hand side, in 142def, 142 denotes abc.
So, abc = 142 and def = 857
Then, abcdef = 142857
So, a + b + c + d + e +f = 1 + 4 + 2 + 8 + 5 +7 = 27
So, the correct answer is “Option B”.
Note: In this question the first thing to remember is we do not have to find all numbers individually but we have to break numbers abcdef and defabc in def and abc only according to their place number so as to get relation on both sides. As the number of digits are 6 so addition must be done correctly as it may change the question.
Complete step-by-step answer:
In this question it is given that there is a six digit number abcdef and when this number is multiplied with 6 then the resulting number is also a six digit number and is defabc. Basically, we have to find the sum of all six numbers that is a + b + c + d + e +f
So, we can denote above statement as
$ 6\cdot (abcdef)=defabc $
As a, b, c, d, e, f are numbers so we can write abcdef as $ 1000\cdot abc+def $ and defabc as $ 1000.def+abc $.
$ 6\cdot (1000\cdot abc+def)=1000.def+abc $
On solving by opening brackets we get
$ 6000\cdot abc+6def=1000.def+abc $
Taking terms of abc on left side and terms of def on right side, we get
$ 6000\cdot abc-abc=1000.def-6def $
On solving, we gte
$ 5999\cdot abc=994\cdot def $
On simplifying coefficients of terms abc and def, we gte
$ 857\cdot abc=142\cdot def $
Now, just see the pattern, what we see is on left side the, in 857abc, 857 denotes def and on right hand side, in 142def, 142 denotes abc.
So, abc = 142 and def = 857
Then, abcdef = 142857
So, a + b + c + d + e +f = 1 + 4 + 2 + 8 + 5 +7 = 27
So, the correct answer is “Option B”.
Note: In this question the first thing to remember is we do not have to find all numbers individually but we have to break numbers abcdef and defabc in def and abc only according to their place number so as to get relation on both sides. As the number of digits are 6 so addition must be done correctly as it may change the question.
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