Question

A six digit number is said to be lucky if the sum of its first three digits is equal to the sum of its last 3 digits. Then the sum of ALL SIX DIGITS lucky number is divisible by
(a.) 2
(b.) 5
(c.) 7
(d.) 13
Please give well explained appropriate answers.
It's a question of PRMO so don't take it easily.
ACCORDING TO THE ANSWER KEY, THE ANSWER IS OPTION (d.) i.e. 13.
So please explain how come!

Answer 1

Answer:

2

Step-by-step explanation:

let sum of first three digits be x

therefore sum of last digits is also equals to x (given)

thus,

sum of all digits = sum of first digits + sum of last digits

= x+x = 2x

Answer 2

Answer:

(a) 2

Step-by-step explanation:

ur expected answer is wrong...bcoz

100100 is a six digit number & sum of first thee digits ( 1 + 0 + 0 ) is equal to that of last three digits ( 1 + 0 + 0 ). but the sum of all six digits ( 1 + 0 + 0 + 1 + 0 + 0 ) is not divisible by 13. but definitely divisible by 2. ( u can try many numbers like this)

actually ..

let the number be ...a b c d e f

a + b + c = d + e + f

now...

a + b + c + d + e + f = 2 ( a + b + c )

obviously it's an even number and...

hence divisible by 2