Hardest Math Problem EVER!!!!! Help!!!?

Question

How many two-digit numbers exist such that when the product of its digits is added to the sum of its digits, the result is equal to the original two-digit number? List the numbers. Explain your reasoning.

Answer 1

ab + a + b = 10a + b
(b + 1) = 10
b = 9
so, we have
19, 29, 39, 49, 59, 69, 79, 89, 99

Answer 2

Let the two digit number be x and y. So the product is x*y and the sum is x+y

So x*y +x+y = 10x +y ; (e.g., if the number is 32, 10(3)+2)
Now subtract y for both

xy + x = 10x
x(y+1)=10x
y+1=10
y=9

So x can be any thing from 1 to 9 but y must be 9 so the answers are

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

So let's check one

49
4*9 + 4+9 =49 ??
36 + 13 = 49
49 = 49 YES!!

Hope this helps!

Answer 3

let's say you have a two digit number, called TO (tens and ones). to make this numerically, we will say there is a digit T and O such that Tx10 + O = TO.

now, what you're asking is what conditions are true when:
T*O + (T+O) = 10T + O

simplifying...

TO + T + O = 10T + O
TO + T = 10T
TO - 9T = 0
T (O-9) = 0

It looks like whenever the ones place is a 9, you will get your answer, as O-9 must equal 0, or T must equal 0. and T cannot equal 0, otherwise it wouldn't be a two digit number. So, any two digit number ending in 9 will work...

Example: 19 = (1*9 + (1 + 9))
19 = 9+10 bingo!

one more ... 49 = (4*9 + (4+9))
49 = 36 + 13
yup!

Answer 4

there are 9 numbers that do this.
19
29
39
49
59
69
79
89
99
Check for yourself

If we define a 2 digit number as YX
Y being the 10s place
X being the 1s place
then any digit can be represented as 10Y+X

Now the digits multiplied together would be Y*X
and the digits added together would be Y+X
so X*Y + (Y+X) = 10Y + X

Subtract X from both sides

X*Y + Y = 10Y

Subtract Y from both sides

X*Y = 9Y

Divide both sides by Y

X = 9

There is only one constrained variable here. As long as it is 2 digits and the last digit is 9, it works.

19
1*9 + 1 + 9 = 19

29
2*9 + 2 + 9 = 29

etc etc.

Answer 5

we call that 2-digits number : x=10a+b
we have ab+(a+b)=10a+b
----->ab=9a
----->a=0 or b=9
if a = 0 --> x=b (wrong bc x is a two-digits number)
if b=9 --->x=10a+9(true)
with a=1,2,3,4,5,6,7,8,9 and b=9 we satify the problem

so the numbers is : 19,29,39,49,59,69,79,89,99

SJSU mathlete

Answer 6

i dont know dude, im a jackass on math seriuosly
=(

long live south park by the way