Let P and S denote the product and the sum of digits of a positive integer.?

Question

How many integers N are there?between 10 and 1000 such that P + S = N

Answer 1

For numbers N between 10-99 = 10A + B
A+B+AB = 10A+B
A(1+B)=10A
B=9
i.e. all numbers ending with 9 (9)

For numbers N between 100 - 999 = 100A + 10B + C
A+B+C + ABC = 100A+10B+C
ABC = 99A + 9B
This means that ABC >= 99A
BC >= 99
since B & C must be less than 10. 9*9 = 81. This is impossible.

Therefore, the number of such integers is 9

Answer 2

Let's try the two-digit numbers of the form 'ab'.

P = a*b
S = a + b
N = 10a + b

So you want:
a*b + a + b = 10a + b

Subtract b from both sides:
a * b + a = 10a

Divide both sides by a:
b + 1 = 10
b = 9

So for the two digit numbers, anything ending in 9 will work:
19, 29, 39, 49, 59, 69, 79, 89 or 99.

Now if you try the same thing for three-digit numbers, you'll find there is no solution.

And obviously 1*0*0*0 + 1+0+0+0 doesn't equal 1000, so this isn't an answer either.

Answer 3

P + S = N

so

P + S = 10 and P + S = 1000

If

P + S = 10, than P = 10 - S

therefore

(10 - S) + S = 1000

10 - S + S = 1000

1000 - 10 = 990

Answer 4

There are 9 such numbers that are 19,29,39,49,59,69,79,89 and 99.

Answer 5

P = 544099
S = 545050
N = 991

Answer 6

between 10 and 1000
sorry i did not know it really
take care

Answer 7

Is there any other formula ?