Question

Lisa wants to use her calculator to square a two-digit positive integer, but she accidentally enters the tens digit incorrectly. When she squares the number entered, the result is 2340 greater than the result she would have gotten had she correctly entered the tens digit. What is the sum of the two-digit number Lisa entered and the two-digit number she meant to enter?

Answer 1

To solve this problem, we can assume the two-digit number Lisa meant to enter as 10a + b and the number she actually entered as 10c + b. By expanding and simplifying the equation, we can find the values of a, b, and c that satisfy the given conditions. The sum of the two-digit number Lisa entered and the number she meant to enter is 131.

Step-by-step explanation:

To solve this problem, let's assume the two-digit number Lisa meant to enter is represented by the variables 10a + b, where a is the tens digit and b is the ones digit. The number Lisa actually entered can be represented as 10c + b, where c is the incorrect tens digit. According to the problem, (10c + b)^2 - (10a + b)^2 = 2340. Expanding and simplifying this equation gives us 100c^2 - 100a^2 + 20cb = 2340. Rearranging the equation and factoring the left side further, we have 20(5c^2 - 5a^2 + cb) = 2340. Dividing both sides by 20, we get 5c^2 - 5a^2 + cb = 117.

Since we're dealing with a two-digit number, a, b, and c can each range from 1 to 9. We can use trial and error to find the values of a, b, and c that satisfy the equation. After testing different combinations, we find that a = 6, b = 7, and c = 4 make the equation true. Therefore, the two-digit number Lisa entered is 64 and the two-digit number she meant to enter is 67. The sum of these two numbers is 64 + 67 = 131.

Answer 2

Answer: Correct two digit number= 24

Step-by-step explanation:

Let

x= tens digit (wrong one)

and y= zeroes digit

and z= tens digit (right one)

Now according to the statement,

(10x+y)² - (10z+y)² = 2340

Simplifying this we get

(x-z)(5x+y+5z)= 117= 1 * 3 * 3 * 13

So it could either be

x-z=1 or x-z=3

If we take the x-z=1

then 5x+y+5z=117

which is not possible as maximum value we can get from this is 99.

This leaves us with x-z=3

so 5x+y+5z= 39

Put x=3+z

We get,

y+10z= 24

Now as y and z both are single digits so the only possible numbers to fit these are y=4 and z=2.

Put z=2 in x=3+z , we get x=5

So the numbers become 10x+y= 54

and 10z+y= 24

Their sum is 54+24=78

Check

54²-24²= 2340 so it is correct.