Question

If abc ,bca,cab are three digit numbers ,and abc+bca+cab=3×(a+b+c)×k then the value of k will be

Answer 1

Answer:

The value of k is 37

Step-by-step explanation:

Any 3 digit number xyz can be written as 100x+10y+z

Now,

abc=100a+10b+c

bca=100b+10c+a

cab=100c+10a+b

Adding these 3 equations, we get

abc+bc+cab=100(a+b+c)+10(a+b+c)+(a+b+c)

abc+bc+cab=(100+10+1)(a+b+c)

abc+bc+cab=(111)(a+b+c)

But given,

abc+bca+cab=3×(a+b+c)×k

⇒ (111)(a+b+c)=3×(a+b+c)×k

⇒ 111=3k

⇒ 3k=111

⇒ k=111/3

⇒ k=37