Find the least number which when divided by 6, 7, 8, 9 and 10 leaves 1, 2, 3, 4 and 5 as remainders respectively, but when divided by 19 leaves no remainder?

A. 8
B. 12
C. 6
D. 2
Let the sides of the rectangle be l and b respectively.
From the given data,
(√l2 + b2) = (1 + 108 1/3 %)lb
=> l2 + b2 = (1 + 325/3 * 1/100)lb
= (1 + 13/12)lb
= 25/12 lb
=> (l2 + b2)/lb = 25/12
12(l2 + b2) = 25lb
Adding 24lb on both sides
12l2 + 12b2 + 24lb = 49lb
12(l2 + b2 + 2lb) = 49lb
but 2(l + b) = 28 => l + b = 14
12(l + b)2 = 49lb
=> 12(14)2 = 49lb
=> lb = 48
Since l + b = 14, l = 8 and b = 6
l – b = 8 – 6 = 2m.