Problem Solving

maihuna wrote: ↑

Tue Mar 15, 2011 12:09 pm

Of the three-digit integers greater than 750, how many have at
least two digits that are equal to each other?
(A) 56
(B) 70
(C) 72
(D) 74
(E) 78

Solution:

In the 700s, if the tens and ones digits are the same, we have: 755, 766, 777, 788, and 799, i.e., a total of 5 such numbers. If the hundreds and tens digits are the same, we have: 770, 771, …, 777, 778 and 779, i.e., a total of 10 such numbers. If the hundreds and ones digits are the same, we have: 757, 767, 777, 787 and 797, i.e., a total of 5 such numbers. It seems there are a total of 20 numbers in the 700s that satisfy the criterion; however, we can see that 777 has been counted three times, so we have to subtract 2 from 20. Thus, the actual number of numbers in the 700s is 18.

In the 800s, if the tens and ones digits are the same, we have: 800, 811, …, 888, and 899, i.e., a total of 10 such numbers. If the hundreds and tens digits are the same, we have 880, 881, …, 888, and 889, i.e., a total of 10 such numbers. If the hundreds and ones digits are the same, we have 808, 818, …, 888, and 898, i.e., a total of 10 such numbers. It seems there are a total of 30 numbers in the 800s that satisfy the criterion; however, we can see that 888 has been counted three times, so we have to subtract 2 from 30. Thus, the actual number of numbers in the 800s is 28.

Like the 800s, there should be another 28 numbers in the 900s. Therefore, there are a total of 18 + 28 + 28 = 74 such numbers.

Alternate Solution:

Notice that there are 999 - 751 + 1 = 249 three-digit integers that are greater than 750. If n is the number of three-digit integers that are greater than 750 such that no two digits of n are the same, then we must have:

The number of three-digit integers greater than 750 where at least two digits are the same = 249 - n

So, let’s calculate n.

In the 700s, the tens digit can be 5, 6, 8, or 9, and the units digit can be any one of the ten digits besides 7 and tens digit. Thus, there are 4 x 8 = 32 choices. However, one of these choices is 750 (which is not greater than 750); therefore there are 32 - 1 = 31 integers greater than 750 where no two digits is the same.

In the 800s, the tens digit can be any digit besides 8, and the units digit can be any digit besides 8 and the tens digit. Thus, there are 9 x 8 = 72 choices.

The above holds also for the 900s; thus, there are 72 integers between 900 and 999 where no two digits are the same as well.

Thus, n = 31 + 72 + 72 = 175. Hence, there are 249 - 175 = 74 three digit integers greater than 750 where at least two of the digits are the same.

Answer: D