Question

Given a positive integer N. Deleting the last digit of N decreases it by 2021. What is N ?

Answer 1

The value of the positive integer N, such that deleting its last digit decreases it by 2021, is 2245. That is, $N=2245$.

Given: A positive integer N such that deleting the last digit of N decreases it by 2021

To find: The value of N

Let x be the integer consisting of all digits of N except the last digit and let the last digit of N be d.

Then, $N=10x+d$

It is given that deleting the last digit  decreases it by 2021

Then, $x=N-2021$

Substitute $N=10x+d$ in $x=N-2021$ to get,

$x=10x+d-2021$

$d=2021-9x$

Since d is a digit, we have,

$0\leq d\leq 9$

Substitute $d=2021-9x$ in $0\leq d\leq 9$ to get,

$0\leq 2021-9x\leq 9$

$9x\leq 2021\leq 9x+9$

$9x\leq 2021\leq 9(x+1)$

$x\leq \frac{2021}{9} \leq x+1$

$x\leq 224.5556 \leq x+1$

Then,

$x\leq 224.5556$ and $224.5556 \leq x+1$

$x\leq 224.5556$ and $x\geq 223.5556$

$223.5556 \leq x \leq 224.5556$

Since x is an integer, the only possibility that satisfies the above inequality is, $x=224$

Substitute $x=224$ in $d=2021-9x$ to get,

$d=2021-9(224)$

$d=2021-2016$

$d=5$

Substitute $x=224$ and $d=5$ in $N=10x+d$ to get,

$N=2245$

Thus, the value of the positive integer N, such that deleting its last digit decreases it by 2021, is 2245. That is, $N=2245$.

Learn more about word problems with digits of a number here:

https://brainly.com/question/17236020