L.C.M. of 5,6,7,8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 X 2 + 3)=1683
Converting given measurements into centimeters 4m 96 cm = 400+96 = 496 cm 4m 3cm = 400+3 =403 cmTo find the side of the largest possible brick , the side of brick must divide length as well as breadth exactly. Using division method 403)496(1 403 93) 403 (4 370 31) 93 (3 93 0HCF = 31 Side of the largest possible square brick =31cm
Let the numbers be 37a and 37b.Then, 37a x 37b = 4107 ab = 3.Now, co-primes with product 3 are (1, 3).So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111). Greater number = 111.
EXAMIANS