Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2004 · Marks: · Difficulty:
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is
(a)4
(b)5
(c)6
(d)8
Answer
Answer (as printed): A
Explanation
$\mathrm{N}=$ H.C.F. of (4665-1305), (6905-4665) and (6905-1305) $$\text { = H.C.F. of 3360, } 2240 \text { and } 5600=1120 .$$ Sum of digits in $\mathrm{N}=(1+1+2+0)=4$.
Explanation as extracted from the printed page; notation may be imperfect.