Let the sum be Rs.P.then P(1+R/100)^3=6690…(i) and P(1+R/100)^6=10035…(ii) On dividing,we get (1+R/100)^3=10025/6690=3/2. Substituting this value in (i),we get: P*(3/2)=6690 or P=(6690*2/3)=4460 Hence,the sum is rs.4460.
How many such pairs of digits are there in the number 421579368 each of which has as many digits between them in the number as when they are arranged in ascending order?