Bunuel
If n, m, and p are distinct, five-digit positive integers and if p = n + m, is the thousands digit of p equal to the sum of the thousands digits of n and m?
(1) The tens digit of p is equal to the sum of the tens digits of n and m.
(2) The hundreds digit of p is equal to the sum of the hundreds digits of n and m.
If n, m, and p are distinct, five-digit positive integers and if p = n + m, is the thousands digit of p equal to the sum of the thousands digits of n and m?
Let n, m & p be expressed as n1n2n3n4n5, m1m2m3m4m5 & p1p2p3p4p5 respectively.
p1p2p3p4p5 = n1n2n3n4n5 + m1m2m3m4m5
Q. Is p2 = n2 + m2 ?
(1) The tens digit of p is equal to the sum of the tens digits of n and m.
p2 = n2 + m2
Since p2 = n2 + m2; there is no carry over of sum of unit digts; p1 = n1 + m1
Since there is no carry over of sum of tens digits;
p3 = n3 + m3; or p3 +1 = n3 + m3
It is not certain whether p2 = n2 + m2
NOT SUFFICIENT
(2) The hundreds digit of p is equal to the sum of the hundreds digits of n and m.
p3 = n3 + m3
There is no carry over of sum of hundredth digit
p2 = n2 + m2
SUFFICIENT
IMO B