Find the sum of all the two digit numbers which can be expressed as an integral multiple of the product of their digits.
I did this by looking at the numbers. It is not as arduous as it looks
10a+b=kab where 1<= a <= 9 0<=b<=9 and k is a positive integer
If a=1 then
10+b=1b * k
b=0 no
b=1 k=11 yes 11
b=2 k=6 yes 12
b=3 no
b=4 14 is not a multiple of 4 so no
b=5 15 is a multple of 5 k= 3 15
b=6 16 is not a multiple of 6 so no
b=7 17 is not a multiple of 7 so no
b=8 18 is not a multiple of 8 so no
b=9 19 is not a multiple of 9 so no
Ok now I can see a bit of a pattern emerging. The original number must be a multiple of the last digit.
so when I look at a=2 I only need think about even 2 digit numbers
10a+b=kab
If a=2 then
20+b= 2b * k
b=2 no
b=4 24 is divisable by 2*4=8 good 24
b=6 no, b=8 no
10a+b=kab a=3
30+b=3b * k
If a=3 I only need consider numbers that are multiples of 3 which are 33,36,39
divide by 3 and I get 11,12,13
these must also be multiples of by b
11/3 no, 12/6 yes 13/9 no
SO only 36 works here
10a+b=kab a=4
40+b=4b * k
If a=4 I only need consider numbers that are multiples of 4 which are 44 and 48
divide by 4 and I get 11 and 12
these must also be multiples of by b
11/4 no 12/8 no
SO none of these work
10a+b=kab a=5
50+b=5b * k
only 55 is a multiple of 5
55/5=11 11 is not a multiple of 5 so no good
10a+b=kab a=6
66 does not work
nor will 77 or 88 or 99
So the only ones that work are 11, 12, 15, 24 and 36.
Guest has already added them for you.