Question

For a positive integer n, let Pn denote the product of the digits of n, and
Sn denote the sum of the digits of n. The number of integers between 10 and
1000 for which Pn + Sn = n is

Answer 1

did you try some numbers

Answer 2

P(10)=1*0 = 0
S(10) = 1 + 0 = 1

P(11) = 1*1 = 1
S(11) = 1 + 1 = 2

Answer 3

so
\(\large n=a_0 10^0 +a_110^1 +a_210^2 +...+a_m 10^m = \sum_{i=0}^m a_i\times 10^i\\
\large P_n=a_0\times a_1 \times a_2...\times a_m=\prod _{i=0}^m a_i \\
\large S_n=a_0+ a_1 + a_2...+ a_m=\sum _{i=0}^m a_i\)

Answer 4

hehe i should try code it :)

Answer 5

yes to program this would work . i dont see an obvious algebraic trick

Answer 6

idk xD i'll see also programming might give a pattern right ?

Answer 7

it might , or just count up the number of answers

Answer 8

\[
\prod _{i=0}^m a_i + \sum _{i=0}^m a_i=\sum_{i=0}^m a_i 10^i
\]

Answer 9

ok then we should expand ?

Answer 10

idk i have just copy pasted that from ur reply abvoe

Answer 11

ik xD
i thought u might have sme idea

Answer 12

this looks more like a program problem as perl said

Answer 13

@dan815

Answer 14

notice that m is fixed here : m=2

Answer 15

they just want the solutions for 3 digit numbers

Answer 16

abc + (a+b+c) = 100a + 10b + c

Answer 17

thats a number theory problem

Answer 18

should we include 1000 as well, we can check that one manually

Answer 19

its false for 1000

Answer 20

ok here is my algorithm
1- set numbers from 10 to 100 in array as n[990]
2- send n[i] to Pn and Sn function
3- if statement to check

Answer 21

So much smart people. XD

Answer 22

abc + (a+b+c) = 100a + 10b + c

99a + 9b = abc
11a + b = abc/9

this is just a diophantine equation in "two" variables
solve the equation for `11a+b = 1` and extrapolate the results

Answer 23

oh lol
abc+a+b+c =100 a +10 b +c

Answer 24

ok lets do it :)

Answer 25

11a + b = 1

by inspection :
a = 1
b = -10
is one solution, so all other solutions can be given by : `a = 1+t`, `b=-10-11t`

Answer 26

next, we can find the values of c, such that `abc/9` is an integer

Answer 27

still wid me ikram

Answer 28

why did you set it equal to 1
11a + b = 1

Answer 29

yepp

Answer 30

that allows us to multiply any number both sides and solve
for exampile, we can multiply abc/9 both sides here

Answer 31

if x,y are solutions of 11a + b = 1
then

x*abc/9 , y*abc/9 will be solutions of 11m + n = abc/9

yeah this is just not good enough

Answer 32

but still give an idea like conditions :O

Answer 33

because we want to solve 11a+b = abc/9
and not 11m+n = abc/9

Answer 34

i got ur point hmm

Answer 35

anyways atleas this much is clear :
9 | abc

Answer 36

remember the divisibility rule for 9 ?

Answer 37

so one of them must be 9 , since they are unit digits

Answer 38

oh right right !

Answer 39

one of them has to be 9

Answer 40

if the sum of digits is divisible by 9

Answer 41

wait , a=9 or b=9 or c=9
or a=3 and b=3 something like this

Answer 42

its a*b*c
not abc, my bad im fully on offtrack on this problem

Answer 43

yup!

Answer 44

9 | a*b*c

Answer 45

thats just an observation, wont help us solve the problem

Answer 46

here is the answer if it helps you revere enginner

```
19
29
39
49
59
69
79
89
99
```

Answer 47

how did u get it ?

Answer 48

program

Answer 49

xD mine dint work ;_;

Answer 50

here is mine
```
foreach my $n (11 .. 999) {
my @digits = split //, $n;
my $sum = eval join '+', @digits;
my $prod = eval join '*', @digits;
print "$n\n" if ($sum+$prod == $n);
}
```

Answer 51

but this problem should be solved using number theory, the solutios are only few and they look pretty patterny

Answer 52

nvm xD
im not good in programming

Answer 53

yeah :)

Answer 54

so basically , 9|abc

Answer 55

9|a*b*c

Answer 56

yeah i now , if im gonna refer to n i would use 100a+10b+c
mathematician xD

Answer 57

okie

Answer 58

from the output, a = 0
any idea why it has to be so ?

Answer 59

so let
ok i might have an idea

Answer 60

9*9*9=729
9+9+9=27
so 729+27=756
so a cant be 9 :)

Answer 61

a=0 gives u 2 digit numbers, so we should consider 2 digit numbers and 3 digit numbers separately hmm

Answer 62

hmm
100a+10b+c = abc + a+b+c
so any digit we would take for a

100a+10b+c > abc + a+b+c

Answer 63

:D
thus we consider a=0

Answer 64

10b+c=bc+b+c

Answer 65

ohk keep going

Answer 66

9b=bc Xd
c=9

Answer 67

wow! that was easy

Answer 68

so all
19
29
39
49
59
69
79
89
99

Answer 69

nice :)

Answer 70

yeah i guess the trick in making a=0

Answer 71

im thinking about this :
100a+10b+c = abc + a+b+c
so any digit we would take for a

100a+10b+c > abc + a+b+c

is there a proof for this

Answer 72

lol i wont note that without seeing output xD

Answer 73

yeah let proof it

Answer 74

99a+9b>ab

Answer 75

claim : the value of a positive integer is always greater than the product of its digits

Answer 76

highest value that ab can take is 9*9 < 100

Answer 77

however the value of n is 10b + c only

Answer 78

:D

Answer 79

no its not clear to me yet

Answer 80

we want to prove

100a + 10b + c > a*b*c

Answer 81

oh lol
ok
100a+10b+c <abc+a+c+b
99a+9b<abc

now highest value that abc could take is 9*9*9 right ?

Answer 82

yes

Answer 83

so in other way we wanna prove this
99a+b > 9*9*9
99*9+9 > 9*9*9 true

Answer 84

wait hmm

Answer 85

100a+10b+c = 10(10a + b) + c > b(ca + b) = abc + b^2 > abc

Answer 86

xD

Answer 87

got it :) that was a very useful trick to stop after 2 digit numbers

Answer 88

wait what u did :O

Answer 89

100a+10b+c = 10(10a + b) + c

Answer 90

fine so far ?

Answer 91

yeah keep going

Answer 92

10 > b and 10 > c so :

100a+10b+c = 10(10a + b) + c
> b(ca + b) + c

Answer 93

fine ?

Answer 94

neat xD
yeah fine

Answer 95

i have replaced 10's with b and c and put the inequality

Answer 96

100a+10b+c = 10(10a + b) + c
> b(ca + b) + c
= abc + b^2 + c
> abc

Answer 97

nishe xD

Answer 98

hehe i tried to show it with maximum bhaha by taking 9 as highest blah
but this is more neat