SOLUTION: The number 35 has the property that when its digits are both increased by 2, and then multiplied, the result is 5 times 7=35, equal to the original number. Find the sum of all two

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Question 1195807: The number 35 has the property that when its digits are both increased by 2, and then multiplied, the result is 5 times 7=35, equal to the original number.
Find the sum of all two-digit number such that when you increase both digits by 2, and then multiply these numbers, the product is equal to the original number.
Found 2 solutions by ankor@dixie-net.com, greenestamps:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
The number 35 has the property that when its digits are both increased by 2, and then multiplied, the result is 5 times 7=35, equal to the original number.
Find the sum of all two-digit number such that when you increase both digits by 2, and then multiply these numbers, the product is equal to the original number.
:
Two digits y & x, the number 10y+x
(y+2) * (x+2) = 10y + x
xy + 2y + 2x + 4 = 10y + x
rewrite this and combine
10y - 2y + x - 2x - xy = 4
8y - x - xy = 4
8y - xy = x + 4
y(8-x) = x + 4
y = %28x%2B4%29%2F%288-x%29
In your graphing calc table, the only single digit integers:
x y
2 | 1, which is: 4*3=12
4 | 2, 6*4=24
5 | 3, 7*5=35
6 | 5, 8*7=56
:
The sum of all these two digit numbers
12 + 24 + 35 + 56 = 127

Answer by greenestamps(13214) About Me  (Show Source):
You can put this solution on YOUR website!


Let x be the tens digit of the 2-digit number and y be the units digit.

Then the number itself is 10x+y.

Adding 2 to each digit and multiplying the two resulting numbers gives the expression (x+2)(y+2).

The conditions of the problem are that

%28x%2B2%29%28y%2B2%29=10x%2By

subject to the constraints that x and y are single-digit integers.

%28x%2B2%29%28y%2B2%29=10x%2By
xy%2B2x%2B2y%2B4=10x%2By

Gather all the terms in one variable on one side of the equation and all terms without that variable on the other, then simplify:

xy%2B2x-10x=y-2y-4
xy-8x=-y-4

Factor out the variable on the left and divide:

x%28y-8%29=-y-4
x=%28-y-4%29%2F%28y-8%29

Perform the division as quotient plus remainder. To do this, given a "-y" term in the numerator and "y-8" as the denominator, rewrite the numerator as a multiple of (y-8) plus a constant; then perform the division:

x=%28%28-y%2B8%29-12%29%2F%28y-8%29
x=%28-y%2B8%29%2F%28y-8%29-12%2F%28y-8%29
x=-1-12%2F%28y-8%29

In this form of the equation, -1 is an integer and x has to be an integer; that means -12%2F%28y-8%29 must be an integer. Note also that, because x must be a positive integer, -12%2F%28y-8%29 must be a positive integer greater than 1. That means y-8 must be negative and a factor of 12.

Try (y-8) equal to each negative number that is a factor of 12 to find solutions:

(1) y-8 = -12 --> y = -4 no; y must be positive
(2) y-8 = -6 --> y = 2; x = -1 + (-12/-6) = 1. The number is 12; (1+2)(2+2) = 3*4 = 12 YES
(3) y-8 = -4 --> y = 4; x = -1 + (-12/-4) = 2. The number is 24; (2+2)(4+2) = 4*6 = 24 YES
(4) y-8 = -3 --> y = 5; x = -1 + (-12/-3) = 3. The number is 35; (3+2)(5+2) = 5*7 = 35 YES
(5) y-8 = -2 --> y = 6; x = -1 + (-12/-2) = 5. The number is 56; (5+2)(6+2) = 7*8 = 56 YES
(6) y-8 = -1 --> y = 7; x = -1 + (-12/-1) = 10 no; x must be a single digit

We have four numbers that satisfy the conditions of the problem: 12, 24, 35, and 56.

ANSWER: 12+24+35+56 = 127