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Find the sum of all the two digit numbers which can be expressed as an integral multiple of the product of their digits.

Dec 8, 2019

#1
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a=1;b=1;p=0; cycle:d=a*10+b;if(d%(a*b)==0, goto loop, goto next); loop:printd,", ",;p=p+n; next:b++;if(b<10, goto cycle, 0);b=1;a++;if(a<10, goto cycle, 0);print"Total = ",p

11 , 12 , 15 , 24 , 36 , Total = 98

Dec 9, 2019
#2
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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.