Skeleton Puzzles 43What number is it that can be multiplied by I, 2, 3, 4, 5, or 6 and no new
figures appear in the result?- ADDING THEIR CUBES
The numbers 407 and 370 have this peculiarity, that they exactly equal the
sum of the cubes of their digits. Thus the cube of 4 is 64, the cube of 0 is 0,
and the cube of7 is 343. Add together 64, 0, and 343, and you get 407. Again,
the cube of 3 (27), added to the cube of 7 (343), is 370. Can you find a
number not containing a zero that will work in the same way? Of course, we
bar the absurd case of I.
- THE SOLITARY SEVEN
*)**(*7***
- ofc
Here is a puzzle, sent me by the Rev. E. F. O. It is the first example I have
seen of one of these missing-figure puzzles in which only one figure is given,
and there appears to be only one possible solution. And, curiously enough, it
is not difficult to reconstruct the simple division sum. For example, as the di-
visor when multiplied by 7 produces only three figures we know the first figure
in the divisor must be I. We can then prove that the first figure in the dividend
must be I; that, in consequence of bringing down together the last two figures
of the dividend, the last but one figure in the quotient must be 0, that the
first and last figures in the quotient must be greater than 7, because they each
produce four figures in the sum, and so on.