Determine the number nearest 110000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Step 1: Find the Least Common Multiple (LCM) of 8, 15, and 21.
Prime factorization:
8 = 2³
15 = 3 × 5
21 = 3 × 7
LCM = 2³ × 3 × 5 × 7 = 8 × 3 × 5 × 7 = 840
So, any number divisible by 8, 15, and 21 must be a multiple of 840.
Step 2: Find the multiple of 840 that is nearest to 110000 and greater than 100000.
Divide 110000 by 840:
110000 ÷ 840 ≈ 130.95
Now check nearby multiples:
130 × 840 = 109200
131 × 840 = 110040
132 × 840 = 110880
Among these, 110040 is the nearest to 110000 and is greater than 100000.
Hence answer is 110040
Real NumbersMore Questions on Real Numbers
In a school there are two sections – section A and section B of class X. There are 32 students in section A and 36 students in section B. Determine the minimum number of books required for their class library so that they can be distributed equally among students of section A or section B.
View solution
Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.
View solution
Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.
View solution
Find the smallest number that when divided by 12, 16, and 24 leaves a remainder of 4 in each case.
View solution
Find the smallest number that is divisible by 6, 10, and 15, and leaves a remainder of 5 in each case.
View solution
If the product of two positive integers is equal to the product of their HCF and LCM is true then, the HCF (32 , 36) is
View solution