Question

let abcdef be a 6 digit integer such that defabc I'd 6 times the value of abcdef. Find the value of a+b+c+d+e+f.


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Answer 1

Answer:

Let’s apply best ever ruling technique developed ; divide and conquer

Break string into 2 parts ABC and DEF and then proceed.

ABCDEF x 7 = DEFABC x 6

[(ABC x 1000) + DEF] x 7 = [(DEF x 1000) + ABC] x 6

7000 ABC + 7 DEF = 6000 DEF + 6 ABC

6994 ABC = 5993 DEF

538 ABC = 461 DEF

538 (461) = 461 (538)

whereby (461) = (ABC) and

(538) = DEF

so.. ABCDEF = 461538

I hope it might help.

Answer 2

Answer:

Sum of the digits of  6 digit integer $(a~b~c~d~e~f)$ is 27.

Step-by-step explanation:

Given a 6 digit integer $(a~b~c~d~e~f)$

And the condition is $6 \times (a~b~c~d~e~f)=(d~e~f~a~b~c)$                     ....(1)

Dividing the number of digits into two sets, i.e, $a~b~c$ and $d~e~f$.

And using the place values

$a~b~c~d~e~f$ can be written as $1000\times a~b~c+d~e~f$

and similarly, $d~e~f~a~b~c~$ can be written as $1000\times d~e~f+a~b~c$

Substituting both in equation (1)

$6 \times (a~b~c~d~e~f)=(d~e~f~a~b~c)$

$\implies 6\big(1000\times a~b~c+d~e~f\big)=1000\times d~e~f+a~b~c$

$\implies 6000~ a~b~c+6~ d~e~f=1000~ d~e~f+a~b~c$

$\implies 6000~ a~b~c-a~b~c=1000~ d~e~f-6~ d~e~f$

$\implies 5999~ a~b~c=994~ d~e~f$

Simplifying the numerical coefficients, i.e., dividing with 7 on both sides,

$857~a~b~c=142~d~e~f$

Therefore, $a~b~c=142~~~\& ~~~d~e~f=857$

Arranging the digits in order, $a~b~c~d~e~f=142857$

And the sum of digits is

$a + b + c + d + e +f = 1 + 4 + 2 + 8 + 5 +7 = 27$

Therefore the value of $a + b + c + d + e +f = 27$