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

First, we have to find the $L.C.M$ of $8, 15$ and$ 21.$
Prime factorization of $8 = 2 \times 2 \times 2$
Prime factorization of $15 = 3 \times 5$
Prime factorization of $21 = 3 \times 7$
Therefore, required $LCM = 2 \times 2 \times 2 \times 3 \times 5 \times 7 = 840$
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 $= 1, 00, 800.$

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