Question

Of the three-digit integers greater than $700$, how many have two digits that are equal to each other and the remaining digit different from the other two?

• A. $90$
• B. $82$
• C. $80$
• D. $45$
• E. $36$

Answer 1

The correct option is C $80$

In three-digit integers, there are three pairs of digits that can be the same while the other digit is different: tens and ones, hundreds and tens, and hundreds and ones. In each of these pairs, there are $9$ options for having the third digit be different from the other two. The single exception to this is in the $700-799$ set, where the number $700$ cannot be included because the problem calls for integers "greater than $700$". So, in the $700-799$ set, there are only $8$ options for when the tens and ones are the same. This is shown in the table below.

Number of digits available for the third digit when two given digits are the same

Same $701-799$ $800-899$ $900-999$

tens and ones $8$ $9$ $9$

hundreds and tens $9$ $9$ $9$

hundreds and ones $9$ $9$ $9$

Thus, of the three-digit integers greater than $700$, there are $9\left(9\right)-1=80$ numbers that have two digits that are equal to each other when the remaining digit is different from these two.

The correct answer is C.