On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders W will be:[/b]  Pick
A) 3, 3, 2
B) 3, 4, 2
C) 5, 4, 3
D) W=5, 5, 2
E) W=6, 4, 3
Let the original number be X. Then by the successive dividing we have followings :   X=5A+2   A=7B+3   B5 +2= 
<gadget id="calculator">5 +2</gadget>
<output>7</output>
7B+3   B=8C+4.   So we have A=7*(8C+4)+3=7*8C+31, and X=5*(7*8C + 31)+2 = 5*7*8C + 157.   Now by dividing X by 8, 7, 5 successively we have followings :   X=8*(5*7C+19)+5   5*7C+19=7*(5C+2) + 5   5C+2=5C+2.   The remainders are, therefore, 5, 5, 2.   The answer is (W=5, 5, 2).
<result>D</result>
On dividing a certain number by 5, 7 and 8 successively, the remainders obtained are 2, 3 and 4 respectively. When the order of division is reversed and the number is successively divided by 8, 7 and 5, the respective remainders W will be:[/b]