This is more difficult then just abc.........

Algebra Level pending

If N is a 6 digit number where 'abcdef' are the digits of N arranged in order, and N = 6 7 × ( d e f a b c ) N= \frac{6}{7} \times (defabc) then find N

538461 461538 676767 767676

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1 solution

Sriharsha Ch
May 23, 2015

a b c d e f abcdef = 6 7 × d e f a b c \frac{6}{7} \times defabc

7 × a b c d e f 7 \times abcdef = 6 × d e f a b c 6 \times defabc

7 × ( 1 0 5 a + 1 0 4 b + 1 0 3 c + 1 0 2 d + 10 e + f 7 \times ( 10^{5}a+10^{4}b+10^{3}c+10^{2}d+10e+f ) = 6 × ( 1 0 5 d + 1 0 4 e + 1 0 3 f + 1 0 2 a + 10 b + c ) 6 \times (10^{5}d+10^{4}e+10^{3}f+10^{2}a+10b+c)

( 7 × 1 0 5 6 × 1 0 2 ) a + ( 7 × 1 0 4 6 × 10 ) b + ( 7 × 1 0 3 6 ) c (7 \times 10^{5} - 6 \times 10^{2})a + (7 \times 10^{4} - 6 \times 10)b + (7 \times 10^{3} - 6)c = ( 6 × 1 0 5 7 × 1 0 2 ) d + ( 6 × 1 0 4 7 × 10 ) e + ( 6 × 1 0 3 7 ) f (6 \times 10^{5} - 7 \times 10^{2})d + (6 \times 10^{4} - 7 \times 10)e + (6 \times 10^{3} - 7)f

1 0 2 × ( 7000 6 ) a + 10 × ( 7000 6 ) b + ( 7000 6 ) c 10^{2} \times (7000-6)a + 10 \times (7000-6)b + (7000-6)c = 1 0 2 × ( 6000 7 ) d + 10 × ( 6000 7 ) e + ( 6000 7 ) f 10^{2} \times (6000-7)d + 10 \times (6000-7)e + (6000-7)f

6994 × ( 1 0 2 a + 10 b + c ) 6994 \times ( 10^{2}a+ 10b +c ) = 5993 × ( 1 0 2 d + 10 e + f ) 5993 \times (10^{2}d + 10e +f )

6994 × ( a b c ) 6994 \times (abc) = 5993 × ( d e f ) 5993 \times (def)

538 ( a b c ) 538(abc) = 461 ( d e f ) 461(def)

As 538 and 461 don't have common factors

a b c = 461 abc=461 a n d and d e f = 538 def=538

N = a b c d e f = 461538 N= abcdef = 461538

Have The options been absent, It would have been a bit tough. The options provide easy trial and error technique to determine the answer

Aditya Narayan Sharma - 5 years, 3 months ago

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good point, actually. thanks for the suggestion!

Sriharsha Ch - 5 years, 3 months ago

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