Find the sum of the digits of the least natural number P, such that the sum of the cubes of the four smallest distinct divisors of P equals 2P.
A.7 B.8 C.9 D. 10
I did like this:-
for 1,2,3,6
2*P=1+8+27+216=252
P=126
all 1, 2, 3 and 6 are factors of 126
therefore digital sum=1+2+6=9 ans
Is there any short method as i took all three cases 1,2,3,4 and 1,2,3,5 and 1,2,3,6.
Best Answer
Let the least number be P, 1 is its least divisor.Let 2nd,3rd and 4th least divisors be x,y and z respectively. We consider the following values of divisor a and the corresponding values of a^3, from x,y and z exactly 1 or all 3 are odd.(P is even) a=1:a^3=1 a=2:a^3=8 a=3:a^3=27 a=4:a^3=64 a=5:a^3=125 a=6:a^3=216 For x, y and z=(2,3,4),2*P=100(i.e.P=50). But 3 is not a divisor of 50. For x,y,z=(2,3,6),2*P=252(i.e.P=126) and the 1,2,3,6 are four least distinct divisor of 126.The required number is 126. The sum of digits is 9.
Do anyone have any other short approach better than i used. it will be of great help. thanks.