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(a) Show that for every nutural number n relatively prime to 10, there is another natural number m all of whose digits are 1 's such that n divides m. (b) Hence or otherwise show that every positive rational number can be expressed in the form (a)/(10^(b)(10^(c)-1)) for some natural a, b, c.

Answer»


Solution :(a) Divide the n+1 number 1, 11, 111, ....,111,....1 (All having only 1 as digits ) by n. Among the n+1 remainders so obtained, two must be equal as the possibilities for remainders are 0, 1, 2,....., n-1 which are n in number. Thus there must be two number x=11.... and y=11.....1 having say j digits and k digits respectively which LEAVE the same remainders after DIVISION by n. We may take `j lt k`. Now we see that y-x is divisible by n. But y-x = 11....100 ... 0 where there are k - j number of 1's and remaining zeros. Since n is COPRIME to 10, we see that n divides m=11 ... 1, a number having only 1's as its digits.
(b) If p/q is any rational number `(p gt 0, q gt0)`, then we may write `q=2^(r)5^(s)t`, where t is coprime to 10.
Choose a number m having only 1's as its digits and is divisible by t. Consider 9m, Which has only 9 as its digits and is still divisible by t. Let k=9 m/t. We see that,
`qk=9m2^(r)5^(s)=(10^(@)-1)2^(r)5^(s)`,
where c is the number of digits in m. Hence we can find d such that `qd=10^(b)(10^(c)-1)` multiply by a suitable power of 2 if `s gt r` and by a suitable power of 5 if `r gt s`). Then
`(p)/(q)=(pq)/(qd)=(a)/(a0^(b)(10^(c)-1))`
where a=pd.


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