Math forum

.any multiple of 3 which is greater than or equal to 6 is possible by combining 9 and 6...as 6 will take care of all the even multiples of 3 and the combination of 9 and 6 will take care of every odd multiple of 3...now in case u order a number which leaves 2 as the remainder on dividing 3 then it will be served with the help of number 20 box..as 20 = 18+2 so when u subtract 20 from tht number then it will give u a number divisible by 3 for eg take 26 it will give u 2 as the remainder on dividing 3 ..now first order 20 number box u r left with 6 which is easily orderable...now we will jump on the case when u get the remainder as 1 on dividing the number by 3...in tht case we will use 2 20 number boxes...now when we subtract 40 from any number with remainder 1 we will surely b left with a number which is a multiple of 3...so at last we came to a conclusion tht any number greater then 40 is orderable provided tht on subtracting 20 or 40 frm it, it turned out to be a multiple of 3 greater then 6 in other words we are sure tht any number greater then 46 is orderable...now we will check for numbers less then 46...45 and 44 are easily orderable as they give 0 and 2 as remainder but wht abt 43?...it will give 1 as a remainder and on subtracting 40 it will give 3 which is not aderable...so 43 is the answer..

thnks madhuri.. :)