Find the squares of the following numbers:

Solution:

Notice that all numbers except the one in question (vii) has 5 as their respective unit digits. We know that the square of a number with the form n5 is a number ending with 25 and has the number n(n + 1) before 25.

(i) Here, n = 42

$\therefore n(n+1)=(42)(43)=1806$

 

$\therefore 425^{2}=180625$

(ii) Here, n = 57

$\therefore n(n+1)=(57)(58)=3306$

 

$\therefore 575^{2}=330625$

(iii) Here n = 40

$\therefore n(n+1)=(40)(41)=1640$

 

$\therefore 405^{2}=164025$

(iv) Here n = 20

$\therefore n(n+1)=(20)(21)=420$

 

$\therefore 205^{2}=42025$

(vi) Here n = 74

$\therefore n(n+1)=(74)(75)=5550$

 

$\therefore 745^{2}=555025$

(vii) We know:

The square of a three-digit number of the form 5ab = (250 + ab)1000 + (ab)2

$\therefore 512^{2}=(250+12) 1000+(12)^{2}=262000+144=262144$

(viii) Here, n = 99

$\therefore n(n+1)=(99)(100)=9900$

 

$\therefore 995^{2}=990025$