Question

do this and help me to come out over this problem​

Answer 1

$\large\sf\underline{Question}$

Find the number nearest to 110000 but greater than 100000 which is exactly divisible by each of 8,15 and 21.

$\large\sf\underline{To\:find}$

• The number nearest to 110000 but greater than 100000 which is divisible by 8 , 15 and 21 .

$\large\sf\underline{Solution}$

The number which is divisible by 8 , 15 and 21 is also divisible by their LCM .

The LCM of 8 , 15 and 21 is :

• $\sf\:8= 2 \times 2 \times 2$

• $\sf\:15= 3 \times 5$

• $\sf\:21= 3 \times 7$

So LCM = $\sf\:2 \times 2 \times 2 \times 3 \times 5 \times 7$

${\sf{{\pink{\:LCM=840}}}}$

Now if we divide 110000 by 840 we will find that it is not completely divisible by 840 and we get 800 as a remainder .

Thus the number nearest to 110000 but greater than 100000 which is exactly divisible by 8 , 15 and 21 and also by 840 ( their LCM ) is :

$\sf\:110000-840$

$\small{\underline{\boxed{\mathrm\pink{➞\:109200}}}}$

!! Hope it helps !!