A person X wants to distribute some pens among six children A B C D E and F. Suppose A gets twice the number of pens received by B, three times that of C, four times that of D, five times that of E and six times that of F. What is the minimum number of pens X should buy so that the number of pens each one gets is an even number?

Now, this is a classical problem, but with a twist pens with A B C D E and F - all should be even. Otherwise, we could have solved by simply taking LCM, ie LCM of 2, 3, 4, 5, and 6 = 60. But now we have to check all values, B C D E F number of pens with B = 60/2 = 30 number of pens with C = 60/3 = 20 number of pens with D = 60/4 = 15 (an odd number) number of pens with E = 60/5 = 12 number of pens with F = 60/6 = 10 Now since adding is not an option, we can simply multiply by 2 convert odd numbers to even. Therefore A becomes 60 × 2 = 120 Similarly doubling all and adding we get total number of pens bought by X = 120 + 60 + 40 + 30 + 24 + 20 = 294