short cut pls!!!

[swapna.rajanish7]

The integer P is greater than 7. If the integer P leaves a reminder of 4 when divided by 9,all of the following must be true except

a.The number that is 4 less than P is a multiple of 9.

b.The multiple that is 5 more than P is a multiple of 9.

c.The number that is 2 more than P is a multiple of 3.

d.When divided by 3,P will leave a remainder of 1.

e.When divided by 2, P will leave a remainder of 1.

Answer is e. Can some one suggest the easiest and the best method to approach this..I dint really follow the explanation. I was wonderin if i can jus plug in numbers and try

[jkandas]

A generic method may be easier...but I was able to quickly narrow it down..just picking first four multiples of 9

I started with 9, 18, 27 and 36. From the question, P leaves a remainder 4 when its divided by 9, so P can be 13, 22, 31 and 40.

Now going through the list, first is pretty much the same as the question stem..all the rest are true except E.

22/2 = 11, remainder =0; 31/2 leaves a remainder of 1. So this is not true.

May be you thought of the same, when you mentioned picking numbers to solve this...

[abhisec]

P can be written as

P = 9q + r (where q is quotient and r remainder.)

therefore,

P = 9q + 4.

a) P-4 is multiple of 9........true put P-4 above you get 9q which is multiple of 9.
b) P + 5= 9q + 9 or 9(q+1) true multiple of 9.
c) P + 2 = 9q + 6 = 3(3q + 2) multiple of 3. true
d) (9q + 4 )/3 ...= (9q + 3 + 1) /3 ......you can see how the remainder is 1. as 9q + 3 is divisible by 3.
e) (9q + 4)/2..............don't know anything about q so cannot calculate the remainder is q =1, remainder can be 1, if q =2 remainder can be 0......so e is the answer.

[Ben Ku]

Please cite the source (author) of this problem. We cannot reply unless a source is cited (and, if no source is cited, we will have to delete the post!). Thanks.