A positive integer X is a six digit number of the form

Question stem:- b=?

St1:- X is divisible by 9
When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9.

2a+4b=9k, where k>1 (\(a\neq0\))

a) a=1, b=4
b) a=9, b=0

So many (a.b) pairs possible.
Insufficient.

St2:- X is divisible by each integer from 1 to 5.

a) when the unit digit of any number is zero, then that number is divisible by at least 1,2 ,and 5.
b) when the last two digits of a number is divisible by 4, then that number is divisible by 4. Since, here the last two digits are 'b'. So '00' is divisible by 4.
c) when the sum of all the digits of an integer is divisible by 3, then the integer is divisible by 3. Now a can be 3 or 6 or 9.
Or, six digit number is divisible by LCM(1,2,3,4,5)=120. So, unit digit has to be 0. Or, b=0.

Insufficient.

Combining, the only possibility of (a.b) is (9,0).

Ans. (C)

Hi PKN

can you please elaborate on statement one, the highlihted part. i didnt get the logic behind it 2a+4b=9k is it some formula for checking divisibility


St1:- X is divisible by 9
When sum of all the digits of an integer is divisible by 9, then the integer is divisible by 9.

2a+4b=9k,where k>1 (\(a\neq0\))

a) a=1, b=4
b) a=9, b=0

Hi dave13,
Question maker has given us a 6-digit integer
Given 6-digit ababbb.

St1 holds when sum of all the digits of the given integer is divisible by 9.

What is the sum of the digits?

Isn't it a+b+a+b+b+b=2a+4b

If I say 'x' is divisible by 'y', then 'x' is a multiple of 'y'. Hope you agree.

Here, 2a+4b has to be divisible by 9. So, (2a+4b) has to be a multiple of 9. Since we don't know the value of multiplying factor, I have assigned it as k.

So, 2a+4b=multiple of 9=9*k

Hope it clarifies your query.

Waiting for further queries(if any).

my approach:
(1)divisible by 9: 2*a+4*b=9m, m is an integer
a b m
1 4 2
2 8 4
insufficient
(2)consider 2,3,4,5
since there're 2 and 5, b must be 0; 3 zeros at the end also means it's divisible by 4
divisible by 3: 2*a divisible by 3 means a is a multiple of 3. a can be 3,6,9; insufficient
1+2: only 9 ensures x is divisble by 9. so a=9