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## #1 Maths Is Fun - Suggestions and Comments » Sides Reversed Is Puzzle » 2013-08-21 03:48:30

SPACKlick
Replies: 1

Error in the solution to this one

Instead of writing 12 x 42, Horace wrote 21 x 24
Both give the answer 504 because 1 x 4 is the same as 2 x 2.
Instead of writing 12 x 84, Horace wrote 21 x 48
Both give the answer 1008 because 1 x 8 is the same as 2 x 4.
We also know that
1 x 8 is the same as 4 x 2,
1 x 6 is the same as 3 x 2,
and 1 x 6 is the same as 2 x 3.
So the other three questions were 14 x 82, 13 x 62 and 12 x 63.

There are many more;
the components missed are
1*4=2*2
1*9=3*3
2*6=3*4
2*8=4*4
2*9=3*6
3*8=4*6
4*9=6*6
The other questions could have been

1) excluding mirror images
13*93=31*39
23*64=32*46
23*96=32*69
24*63=42*36
24*84=42*48
26*93=62*39
34*86=43*68
36*84=63*48
46*96=64*69

2) excluding all the same digit ones
12*21=21*12
13*31=31*13
14*41=41*14
15*51=51*15
16*61=61*16
17*71=71*17
18*81=81*18
19*91=91*19
23*32=32*23
24*42=42*24
25*52=52*25
26*62=62*26
27*72=72*27
28*82=82*28
29*92=92*29
34*43=43*34
35*53=53*35
36*63=63*36
37*73=73*37
38*83=83*38
39*93=93*39
45*54=54*45
46*64=64*46
47*74=74*47
48*84=84*48
49*94=94*49
56*65=65*56
57*75=75*57
58*85=85*58
59*95=95*59
67*76=76*67
68*86=86*68
69*96=96*69
78*87=87*78
79*97=97*79
89*98=98*89

3) excluding quadruple digits
11*22=11*22
11*33=11*33
11*44=11*44
11*55=11*55
11*66=11*66
11*77=11*77
11*88=11*88
11*99=11*99
22*33=22*33
22*44=22*44
22*55=22*55
22*66=22*66
22*77=22*77
22*88=22*88
22*99=22*99
33*44=33*44
33*55=33*55
33*66=33*66
33*77=33*77
33*88=33*88
33*99=33*99
44*55=44*55
44*66=44*66
44*77=44*77
44*88=44*88
44*99=44*99
55*66=55*66
55*77=55*77
55*88=55*88
55*99=55*99
66*77=66*77
66*88=66*88
66*99=66*99
77*88=77*88
77*99=77*99
88*99=88*99

11*11=11*11
22*22=22*22
33*33=33*33
44*44=44*44
55*55=55*55
66*66=66*66
77*77=77*77
88*88=88*88
99*99=99*99

## #2 Maths Is Fun - Suggestions and Comments » Absolutely Christmas Crackers Puzzle » 2013-08-21 01:46:52

SPACKlick
Replies: 2

There is an error on this page

3 x 4 x 4 = 48. Sadly at least two crackers must be exact duplicates of ones already made.

BUT if you could handle the sad look on people's faces, you could consider a missing item as a possible combination. This gives 4 x 5 x 5 = 125 different crackers. Many crackers would be missing one item, some missing two, and one cracker with nothing in at all! (This solution by Eric Chen.)

4x5x5=100 not 125

Also rather than Many and Some why not say 40 crackers will be missing 1 item and 14 will be missing 2?

23894

(2^3)-8+9+4=13