In base ##7##, only the digits ##0,1,2,3,4,5,6## can be used. So ##7 \times 7 = 49## in base ##10##, but ##49## is a multiple of ##7## seven times, so at the right-most digit position of the answer, I can put down a ##0## and carry a ##7 ## to the next-left digit position. I can't put down ##7##...