Playing with Numbers Test - 2

X leaves a remainder 4, when divided by 5.
(X - 4) is divisible by 5.
The ones digit of X - 4 is either 0 or 5.
So, the ones digit of X is either 4 or 9.

To determine if a number is divisible by 11, we need to find the difference between the sum of the digits at even places and the sum of the digits at odd places.
If that difference is divisible by 11, then the number is also divisible by 11.
So, in 383X:
8 + X - 3 - 3 = 0 or Multiple of 11

(i) 8 + X - 6 = 0
X + 2 = 0
X = -2 (Negative number not possible)

(ii) 8 + X - 6 = 11
X + 2 = 11
X = 9

Putting X = 9, we get
8 + 9 - 3 - 3 = 11
Since 11 is divisible by 11, so the number is 3839.

The three-digit number is bca.
Sum of its digits = a + b + c

The rule of 3 states that if the sum of the digits of a number is divisible by 3, then the number will also be divisible by 3.
Therefore, if a + b + c is divisible by 3, then 'bca' will be divisible by 3.

Divisor = (6 × 45) = 270
10 × Quotient = 270
Quotient = 270 ÷ 10 = 27

Dividend = (Divisor × Quotient) + Remainder
= (270 × 27) + 45
= 7290 + 45
= 7335

The Rule for 6: The prime factors of 6 are 2 and 3.

So, for a number to be divisible by 6, it must also be divisible by 2 and 3.
Therefore, we need to check if a number is even and then check if the sum of the digits is divisible by 3.

If a number is divisible by 2 and 3, then it is divisible by 6 only.
So, statement A is incorrect.

(I) Let ab be the required number and number obtained by reversing the digits is ba.
So, ab + ba = (10a + b) + (10b + a) = 11a + 11b = 11(a + b). So, it is divisible by 11.

(II) The Rule for 6: The prime factors of 6 are 2 and 3. So, for a number to be divisible by 6, it must also be divisible by 2 and 3. Therefore, we need to check if a number is even and then check if the sum of the digits is divisible by 3. The number is divisible by both 2 and 3. Since, PQR is divisible by 6, it is also divisible by 2 and 3.

(III) Let ab be the required number and number obtained by reversing the digits is ba.
So, ab - ba = (10a + b) - (10b + a) = 9a - 9b = 9(a - b) it is divisible by 9. Therefore, Z = 9.

(A) Given number: 85235x
For divisibility by 11, (8 + 2 + 5) - (5 + 3 + x) = 0 or Multiple of 11
15 - 8 - x = 0, 11
x = 7 (Only possible value)

(B) Given number: 65x92
For divisibility by 6, last digit should be even and the sum of the digits should be divisible by 3.
So, 6 + 5 + x + 9 + 2 = Multiple of 3
22 + x = Multiple of 3
So, x can be 2, 5 or 8.

(C) Given number: 78430x
For divisibility by 8, the number formed by the last three digits of the number should be divisible by 8.

So, the only possible value of 'x' such that 30x is divisible by 8 is 4.

(D) Given number: 7869x2
For divisibility by both 4 and 3, the number formed by the last two digits of the number should be divisible by 4, and the sum of the digits should be divisible by 3.
So, the values of 'x2' in 7869x2, such that it is divisible by 4, can be 12,32,52,72 and 92.

Now, for the number to be also divisible by 3, the sum of the last two digits of the number should also be divisible by 3.(As sum of the remaining digits is divisible by 3)
Among 12, 32, 52, 72 and 92, it only holds true for 12 and 72.
So, x can be either 1 or 7.

For divisibility by 6, sum of digits has to be a multiple of 3 and the last digit should be even.
A number is divisible by 3, if the sum of the digits is divisible by 3.
5 + X + X + X + X = 4X + 5
4X + 5 should be divisible by 3 and X should be an even number.

Put X = 0.
4X + 5 = 5, which is not divisible by 3.

Put X = 2.
4X + 5 = 13, which is not divisible by 3.

Put X = 4.
4X + 5 = 21, which is divisible by 3.
So, X = 4