We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.More informationAgree
ad
profilesara
asked 2017-01-18 06:50:27 
how many four digit numbers begin and end with an even numbers? Can we solve this problem through arithmetic progression? how?
like 0 dislike 0 comment
profilejiju
answered 2017-01-21 18:21:53 
Ans: $2000$

Solution 1: Using Arithmetic Progression


Consider the numbers $2000,2002,\cdots,2998$
Count of these numbers
$=\dfrac{2998-2000}{2}+1=500$

Similarly,  each of $(4000,\cdots,4998),$ $(6000,\cdots,6998),$ $(8000,\cdots,8998)$ gives $500$ numbers.

So, required count
$500+500+500+500=2000$

Solution 2 (using combinatorics)

First digit can be any of $(2,4,6,8)$
Second digit can be any of  the $10$ digits
Third digit can be any of  the $10$ digits
Fourth digit can be any of $(0,2,4,6,8)$

Required count
$=4×10×10×5=2000$
like 0 dislike 0 comment

Answer This Question

?
Name
    cancel
    preview