Divisibility

In order for 6587126X4 to be divisible by 8, the number formed by the last three digits should be divisible by 8. 

∴ 6X4 must be divisible by 8.

∴ X must be equal to 2 or 6. 

Hence, option (c).

As the divisor is 18, which is not a prime number, we need to write it as product of co-prime number.

i.e. 18 = 2 × 9.

For the number to be divisible by 18, it should be divisible by both 2 and 9.

For 2: Unit place should be divisible by 2.

Since, last digit is 2, it is divisible by 2.

For 9: Sum of digits of the number should be divisible by 9.

∴ k + 6 + 5 + 4 + 3 + 2 = 20 + k should be divisible by 9.

As k is a single digit number, then the only value k can take is 7

Hence, 7.

In order for x754 to be divisible by 11, 

Sum of digits at odd places (o) = 7 + 4 

Sum of digits at even places (e) = x + 5

For this number to be divisible by 11, difference of ‘o’ and ‘e’ should be divisible by 11.

∴ o – e = (7 + 4) – (x + 5) = (6 – x) should be divisible by 11.

⇒ x = 6

In order y246 to be divisible by 9, sum of its digits should be divisible by 9.

∴ y + 2 + 4 + 6 = y + 12 should be divisible by 9. 

The only number which satisfies this condition is 6. 

Hence y = 6

∴ xy = 66

Hence, option (b).

As divisor is 72, then given number must be divisible by 8 & 9 both.

For 9: sum of digits of the number should be divisible by 9.

4 + 2 + 5 + 7 + 3 + k = 21 + k, for the number to be divisible by 9 value k can take 6

21 + k = 21 + 6 = 27 as 27 is divisible by 9, then 425736 is also divisible by 9.

For 8: Last 3 digit of the number should be divisible by 8.

If we take 6 as in unit place then, 736 is last 3 digits then 736 is divisible by 8.

Therefore k = 6.

Hence, 6.

For a number to be divisible by 3, the sum of the digits should be divisible by 3.

Here sum of the digits = 1 + 2 + 2 + 5 + x + 2 + 4 = 16 + x.
16 + x is divisible by 3 when x = 2 or 5 or 8.

The number can be 1225224 or 1225524 or 1225824

For a number to be divisible by 8, last 3 digits should be divisible by 8.
Both 224 and 824 are divisible by 8, hence 1225224 and 1225824 are also divisible by 8.

∴ 1225224 and 1225824 are divisible by both 2 and 8.

Hence, option (d).

Co-prime factors of 80 are 5 and 16. If the number is divisible by both 5 and 16, we can say that the given number is divisible by 80.

Since the number is divisible by 16, it has to be an even number which is also divisible by 5. This is possible only when last digit is 0. Hence, y = 0

The number is 653x0.

This number is divisible by 5.
To check if it is divisible by 16, we have to consider the last four digits of the number.

Option (a): Let x = 2
Last four digits  =  5320 (not divisible by 16).

Option (b): Let x = 4
Last four digits  =  5340 (not divisible by 16).

Option (b): Let x = 5
Last four digits  =  5350 (not divisible by 16).

Option (b): Let x = 6
Last four digits  =  5360 (divisible by 16).
∴ x + y = 6 + 0 = 6

Hence option (d).

Option (a): For a number to be divisible by 2, the last digit should be divisible by 2.
Since the last digit of the given number is no divisible by 2, the number is also not divisible by 2.

Option (b): For a number to be divisible by 3, sum of its digits should be divisible by 3.
Sum of digits = 6 + 7 + 2 + 1 = 16 which is not divisible by 3, hence the number is also not divisible by 3.

Option (c): For a number to be divisible by 5, the last digit should be divisible by 5.
Since the last digit of the given number is no divisible by 5, the number is also not divisible by 5.

Option (d): For a number to be divisible by 11, the difference of sum of odd digits and sum of even digits should be divisible by 11.
Sum of odd digits (o) = 7 + 1 = 8
Sum of even digits (e) = 6 + 2 = 8
o - e = 0, which is divisible by 11, hence the number is also divisible by 11.

Hence, option (d).

If a number is divisible by a and b, it is also divisible by LCM(a, b).

Option (a): A number which is divisible by 12 and 3 is also divisible by LCM(12, 3) = 12.
∴ It cannot be definitely said that the number is divisible by 36.

Option (b): A number which is divisible by 4 and 9 is also divisible by LCM(4, 9) = 36.
∴ Option (b) is definitely true.

Option (c): A number which is divisible by 12 and 9 is also divisible by LCM(12, 9) = 36.
∴ Option (c) is definitely true.

Hence, option (d).

If 146x is divisible by 7, then 156 – 2x should also be divisible by 7.

∴ 146 - 2x should be either 140 or 133 (divisible by 7)

Case 1: 146 - 2x = 140
⇒ x = 3
∴ The number is 1463.
But this number is not divisible by 4.

Case 2: 146 - 2x = 133
⇒ x = 13/2
Not possible since x has to be a single digit integer.

Hence, option (d).