Question 15 Marks
Determine the number nearest to 100000 but greater than 100000 which is exactly divisible by each of 8, 15 and 21.
Answer
View full question & answer→First, we have to find the L.C.M of 8, 15 and 21.
Prime factorization of 8 = 2 × 2 × 2
Prime factorization of 15 = 3 × 5
Prime factorization of 21 = 3 × 7
Therefore, required LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840
The number nearest to 1, 00,000 and exactly divisible by each 8, 15 and 21 should also be divisible by their LCM (i.e. 840)
We have to divide 1, 00,000 by 840.
Remainder = 40
Therefore, Number greater than 1, 00, 000 and exactly divisible by 840 = 1, 00, 000 + (840 - 40) = 1, 00, 000 + 800 = 1, 00, 800
Therefore, Required number = 1, 00, 800.
Prime factorization of 8 = 2 × 2 × 2
Prime factorization of 15 = 3 × 5
Prime factorization of 21 = 3 × 7
Therefore, required LCM = 2 × 2 × 2 × 3 × 5 × 7 = 840
The number nearest to 1, 00,000 and exactly divisible by each 8, 15 and 21 should also be divisible by their LCM (i.e. 840)
We have to divide 1, 00,000 by 840.
Remainder = 40
Therefore, Number greater than 1, 00, 000 and exactly divisible by 840 = 1, 00, 000 + (840 - 40) = 1, 00, 000 + 800 = 1, 00, 800
Therefore, Required number = 1, 00, 800.