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

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.