let n be the greatest number that will divide 1305 , 4665 and 6905 , leaving the game remainder in each case . then sum of the digits in n is :
solution n = h . c . f . of ( 4665 - 1305 ) , ( 6905 - 4665 ) and ( 6905 - 1305 ) = h . c . f . of 3360 , 2240 and 5600 = 1120 . sum of digits in n = ( 1 + 1 + 2 + 0 ) = 4 . answer a
add(add(const_1, const_1), const_2)