The repeating decimals 0. a b a b a b ¯ and . a b c a b c a b c ¯ satisfy 0. a b a b a b ¯ + 0. a b c a b c a b c ¯ = 33 37 where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.

Question

The repeating decimals   0.  a  b  a  b            a      b        ¯   and   .  a  b  c  a  b  c            a      b      c        ¯   satisfy   0.  a  b  a  b            a      b        ¯    +  0.  a  b  c  a  b  c            a      b      c        ¯    =      33    37  where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.

vyhlodatis · Accepted Answer

Step 1Let   x  =  0.            a      b        ¯   and   y  =  0.            a      b      c        ¯   . Then   100  x  =  x  +  a  b , so   x  =            a      b        99   , and, similarly,   y  =            a      b      c        999   . You now know that            a      b        99    +            a      b      c        999    =      33    37      .Put everything over the common denominator of 10989, and you find that  (  111  )  (  a  b  )  +  (  11  )  (  a  b  c  )  =  9801    .Now   (  111  )  (  a  b  )  =  a  b  00  +  a  b  0  +  a  b , and   (  11  )  (  a  b  c  )  =  a  b  c  0  +  a  b  c , so   a  +  a  +  possible carry  =  9 . Clearly there is a carry, and   a  =  4 . See if you can finish it from there.An alternative (and definitely easier) approach is to use the same idea to notice that   x  =            a      b      a      b      a      b        999999    and   y  =            a      b      c      a      b      c        999999   , so that  a  b  a  b  a  b  +  a  b  c  a  b  c  =  891891    .From here it’s easy to get   a  =  4 , and the other two letters then fall into places quite easily as well.

HypeMyday3m · Accepted Answer

Step 1The numbers can be expressed as   0.            a      b        ¯   and   0.            a      b      c        ¯   and written as fractions fractions as       0.                  a        b            ¯        =                    a        b            99       and   0.            a      b      c        ¯    =                    a        b        c            999       :                    a        b            99        +                    a        b        c            999        =            33      37      Let   h  =  a  b and   k  =  c    ⟹    a  b  c  =  10  h  +  kwhere   0  ≤  h  &amp;lt;  100 and   0  ≤  k  &amp;lt;  10 .Rewriting the same expression using h and k :            h      99        +                    10        h        +        k            999        =            33      37      We notice that   999  =  37  ×      3    3   and   99  =      3    2    ×  11            h                        3          2                ×        11              +                    10        h        +        k                    37        ×                  3          3                      =                    33        ×                  3          3                            37        ×                  3          3                    multiply by       3    2   :            h      11        +                    10        h        +        k                    37        ×        3              =                    11        ×                  3          3                    37      Since h and k are natural numbers, the only way this expression can be satisfied is by having h to be divisible by 11 and   10  h  +  k to be divisible by 3.let   h  =  11  n where   n  ∈  {  0  ,  1  ,  2  ,  3  ,  4  ,  5  ,  6  ,  7  ,  8  ,  9  }so      n    +                    10        ×        11        n        +        k                    37        ×        3              =                    11        ×                  3          3                    37      some algebra and we get  221  n  +  k  =  891since   k  &amp;lt;  10 and   221  ×  4  =  884 we see that   n  =  4 gets us close.  884  +  k  =  891    ⟹    k  =  7putting it all together:  c  =  7  a  b  =  h  =  11  ×  4  =  44    ⟹    a  =  4  ,  b  =  4

The repeating decimals 0.abab overline(ab) and .abcabc overline(abc) satisfy 0.abab overline(ab)+0.abcabc overline(abc)=3337 where a,b, and c are (not necessarily distinct) digits. Find the three-digit number abc.

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