text: framed? haha? kevin had five rectangular photographs. the ten side lengths of the photographs were 1cm,. 2cm, 3cm... 10cm (with each occuring once).  katie has five rectangular photographs with side lengths 1cm., 2cfm, 3cm... 10cm (with each occuring once). she assembled hers into a rectangular frame of size 15cm * 10cm. find the side lengths of each photobgraph and show how the photographs can be put in teh frame. let us assume that each photograph is rectangular. the photographs must have different length of sides, which are in the range 1 through 10. also, the sides of the photographs must be such that the sum of the area of all of them equals 150 cm^2. further, the sides should be such that these 5 photographs can be arranged in a 15 cm * 10 cm frame. if the dimensions of the n-th photograph be ‘an’ and ‘bn’ then  a1*b1 + a2*b2 + a3*b3 + a4*b4 + a5*b5 = 150 _________(1).  by trial we find 10*1 + 9*8 + 7*6 + 5*4 + 3*2 = 10 + 72 + 42 + 20 + 6 = 150.  we also find that these dimensions can be arranged in a rectangle of dimensions 15*10 as shown in the fig. that you i have sent you by email.  the form 10*1 implies a rectangle which has its 10 unit long side parallel to the x-axis and 1 unit long side parallel to the y-axis.  however, not all the combinations satisfying equation (1) can be arranged in the aforesaid frame. let us take the combination [1*8, 2*7, 3*6, 4*5, 9*10]. the sum of these 5 terms is 150 but these 5 rectangles cannot be arranged in a 15*10 frame.