Question

$199988,1704382,200175,1702497,201200,1712040$

Verified by Toppr

Hence $1712040$ is the largest number and the digits in the remaining two numbers at thousands place are $4$ and $2$

Since $4>2$

Therefore $1704382>1702497$

$201200,200175$ and $199988$ are $6−$digits numbers in which $2$ and $1$ are at lakhs place respectively

Since $2$ is greater than $1$

Therefore $201200>200175$

Thus $199988$ is the smallest number

The given numbers in the descending order are:

$1712040>1704382>1702497>201200>200175>199988$

0

0