Question

How many 2-digit numbers are there which are divisible by the sum of their digits ?

Answer 1

Is it supposed to be an algorithm? Or your maths teacher is a kamikazi?

Answer 2

no odd numbers have this property ( i guess!}

Answer 3

I mean.. This is quite easy to do with a program but ffs trying it by hand its really hard

Answer 4

Diogo - Its just the count of such two digit numbers :)

Answer 5

i'i wrong - 63 is divisible by 9

Answer 6

It's true for all products of 9.

Answer 7

Except 99.

Answer 8

right!

Answer 9

yes indian, but first you need to find them all.. anyway, the sum of any number that is a multiple of 3 is divisible by 3.

example: 15,21,24,27,30,33,36,39,42, etc

Answer 10

99=3+(n-1)*3

Answer 11

find n=??? that will your answer

Answer 12

15 must be divisible by 1+5 = 6

Answer 13

33

Answer 14

ohhh. you're right mkuma36. I didnt understand the question at first. So 20 is divisible by 2, 30 by 3, 40 by 4, 50 by 5, 60 by 6, 70 by 7, 80 by 8, 90 by 9

Answer 15

Exactly Diogo. But 33 is not the right answer I guess

Answer 16

only 21 and 81 fit the bill for x1 up to 100

Answer 17

yes Jimmyrep that is corect

Answer 18

only 63 for x3 up to 100

Answer 19

For x5, the only one that works is 50.

Answer 20

right

Answer 21

this can be done by counting - not as tedious as i thought

Answer 22

Even for the counting you can use math. Reduce the question into an eqn with two variables (the digits). It helps.

Answer 23

33 also works for x3

Answer 24

there are n x7 that works

Answer 25

No dalvy.. 33 is not divisible by 6

Answer 26

no x7

Answer 27

I am silly.

Answer 28

lol

Answer 29

only 18 and 49 for x8

Answer 30

lol 48!!!!!

Answer 31

how it is 48??

Answer 32

48 is divisible by 12.

Answer 33

there are some weird results in number theory. for example the fraction of numbers between 1-100 which have two or more repetetive factors is 6 / pi^2!
H ow the heck can pi be involved in something like that?

Answer 34

Ok so I just did a visual basic program to solve this:

here is the algorithm:
Dim i, numero As Integer
Dim j, a, k As Double
For i = 10 To 99
j = i / 10
a = Math.Truncate(j)
k = j - a
k = k * 10
numero = k + a
If i Mod numero = 0 Then
MsgBox(i)
End If
Next
-------------------------------------------
The values I found are: 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90

Thats all.

Good luck

Answer 35

We know the divisibility rule of 3.
It says that if the sum of digits are divisible by 3 or 9 then the number is divisible by 3 or 9.
Can't be take the series as 3,6,9,12,15...

Answer 36

i got 23 but it might wrong

Answer 37

oh diogo got that many too
i went a long process
i wrote down every sum possible from 0 to 9
and check both digits and each sum to see if that number would be divisible by the sum
and i ended up counting 23

Answer 38

myininaya - you are correct and so is Diogo.. Good job mates. Diogo going out of the way to solve this problem... woohoo!!

Answer 39

he actually has an algorithm up there

Answer 40

Cool. Here is the short cut or so I think - (10a+b)/a+b=positive integer
a+b+9a/a+b
so we just hv 2 make 9a/a+b an positive integer
value of A vary from 1 to 9
value of B vary from 0 to 9
so just put d values n u ll strt getting d reqd. numbers

numbers are
10 12,21,24,42,48,84
20 multiples of 9=18,27,36,45,54,63,72,81
30
40
50
60
70
80
90

Answer 41

thats cool

Answer 42

myininaya, Diogo actually wrote a computer program to solve this.. Max respects!