1. ## fenders

choose a number say 156
its factors are 1,2,3,4,6,12,13,26,39,52,156
grouped by last digit (1) (2,12,52) (3,13) (4) (6,26,156) (78) (39)

We say the FENDERS (factor enders) of 156 are 1,2,3,4,6,8,9 &that 156 is a 7-fender (it has seven fenders)
A) Show that a number which has 0 & 9 as fenders has at least four more fenders.
B) Find 3 9-fenders less than 1000 with different sets of fenders?
I know 900 is one(0,1,2,3,4,5,6,8,9) & 420 is another (0,1,2,3,4,5,6,7,8) but can't find another?

2. This strange and agruably pointless problem has already been answered here .

Also...

(B) 630 (0,1,2,3,5,6,7,8,9)

3. ## fenders

Originally Posted by Quick
This strange and agruably pointless problem has already been answered here .

Also...

(B) 630 (0,1,2,3,5,6,7,8,9)
630 is a 10 fender (45*14) so still need help

4. Originally Posted by Quick
This strange and agruably pointless problem has already been answered here .

Also...

(B) 630 (0,1,2,3,5,6,7,8,9)
Help, how 630 is a 8 fender?

KeepSmiling
Malay

5. Originally Posted by malaygoel
Help, how 630 is a 8 fender?

KeepSmiling
Malay
35*18 gives 630

6. Originally Posted by bardo
B) Find 3 9-fenders less than 1000 with different sets of fenders?
I know 900 is one(0,1,2,3,4,5,6,8,9) & 420 is another (0,1,2,3,4,5,6,7,8) but can't find another?
It is 270(0,1,2,3,5,6,7,8,9)
KeepSmiling
Malay

7. Originally Posted by malaygoel
It is 270(0,1,2,3,5,6,7,8,9)
KeepSmiling
Malay
Nope...$\displaystyle 54\times5=270$

There must be some sort of method we are missing, surely a teacher wouldn't make you guess and pick.

8. Originally Posted by Quick
Nope...$\displaystyle 54\times5=270$

There must be some sort of method we are missing, surely a teacher wouldn't make you guess and pick.
980 is the other 9 fender with 0,1,2,4,5,6,7,8,9