Divisibility Rules

[harsh.champ]

The product of two integers is 27*33*55*73.Then the sum of the two numbers may be divisible by

(A)16
(B)9
(C)25
(D)49
(E)18

The OA is C.

[shashank.ism]

The product of two integers is 27×33×55×73.Then the sum of the two numbers may be divisible by

(A)16
(B)9
(C)25
(D)49
(E)18

The OA is C.

If a number d divides a number a, but does not divide another number b, then d does not divide a+ b (or a - b).
Here, the product of the two numbers, say A and B, is equal to27×33×55×73.
If A = 2mM and B=2nN where M and N are not divisible by 2, then m + n = 7. So either m>=4 or n>=4,but not both. Thus, we see that either A or B, but not both, is divisible by 24 = 16.

Similarly, we find that either A and B, but not both is divisible by 32 = 9.The same is true with 72 instead of 32.However both A and B can be divisible by 52 = 25.For example, A =52 and B = 27×33×55×73.So the correct choice is option (3)

[ajith]

The product of two integers is 27*33*55*73.Then the sum of the two numbers may be divisible by

(A)16
(B)9
(C)25
(D)49
(E)18

The OA is C.

May be or may not be?

It should not be divisible by 25 for any combination

And I can see combinations which are divisible by 9

[ajith]

The product of two integers is 27×33×55×73.Then the sum of the two numbers may be divisible by

(A)16
(B)9
(C)25
(D)49
(E)18


Similarly, we find that either A and B, but not both is divisible by 32 = 9.The same is true with 72 instead of 32.However both A and B can be divisible by 52 = 25.For example, A =52 and B = 27×33×55×73.So the correct choice is option (3)

what about 9 and 3*33*55*73= 9 and 397485 the sum is also divisible by 9