Of the three digit numbers greater than 500, how many have exactly one

Question

Of the three digit numbers greater than 500, how many have exactly one digit repeating?

a) 134
b) 135
c) 136
d) 142
e) 148

Source: Expert's global practice tests

Bunuel · Accepted Answer

Three-digit number can have only following 3 patterns: A. all digits are distinct; B. two digits are alike and third is different; C. all three digits are alike. We need to calculate B. B = Total - A - C The number of three-digit numbers which are greater than 500 is 499 (from 501 to 999, inclusive). A. all digits are distinct = 5*9*8 = 360 (first digit can have only five values 5, 6, 7, 8, or 9); C. all three are alike = 5 (555, 666, 777, 888, 999). So, 499 - 360 - 5 = 134. Answer: A. Similar questions: https://gmatclub.com/forum/of-the-three ... 28853.html https://gmatclub.com/forum/of-the-three ... 27390.html https://gmatclub.com/forum/of-the-three ... 35188.html Hope it helps.

lifeforhuskar · Answer

i did it in a different way
so whatever the number will be , it will be (multiple of 5) -1
because in each series we will have 600, 700, 800,900 but not 1000, 
only 134 is that kind of number

did it in 30 seconds

Hit kudos if u like it

Trev50 · Answer

This won't work if all options were in the form 5n -1

Posted from my mobile device

gracie · Answer

three ways: 
xxy
xyx
yxx
3(5*9*1)=135
135-1 (for 500)=134
A

800Dreamer · Answer

hi, Is there any way which doesn't subtract unfavorable cases and directly count favorable cases? I know the above one is the best efficient but want to know other possibility as well.

Kinshook · Answer

Asked: Of the three digit numbers greater than 500, how many have exactly one digit repeating?

Numbers can be of form = {abb, bab, bba}

Form abb;
a = {6,7,8,9} = 4 ways
b = {0,1,2,3,4,5,6,7,8,9} except digit a = 9 ways
Total numbers = 4*9 = 36 numbers
Let us take a = 5
b = {1,2,3,4,6,7,8,9} = 8 ways
Total numbers = 36 + 8 = 44 numbers

Form bab;
b = {5,6,7,8,9} = 5 ways
a = {0,1,2,3,4,5,6,7,8,9} except digit b = 9 ways
Total numbers = 5*9 = 45 numbers

Form bba;
b = {5,6,7,8,9} = 5 ways
a = {0,1,2,3,4,5,6,7,8,9} except digit b = 9 ways
Total numbers = 5*9 = 45 numbers

Adding all cases = 44 + 45 + 45 = 134 numbers

IMO A

Kinshook · Answer

nkhl.goyal 
Please see my solution and suggest any changes.

800Dreamer · Answer

hi, Is there any way which doesn't subtract unfavorable cases and directly count favorable cases? I know the above one is the best efficient but want to know other possibility as well.

nkhl.goyal 
Please see my solution and suggest any changes.

Yes, I was looking the same.

For 1st part : 
Form abb;
a = {6,7,8,9} = 4 ways
b = {0,1,2,3,4,5,6,7,8,9} except digit a = 9 ways
Total numbers = 4*9 = 36 numbers
Let us take a = 5
b = {1,2,3,4,6,7,8,9} = 8 ways
Total numbers = 36 + 8 = 44 numbers

You can directly do this. a = {5, 6,7,8,9} = 5 ways, b same as yours.
Total numbers = 5*9 = 45, but here 500 is included so remove that. So, total 45-1 = 44

MathRevolution · Answer

Three digit numbers greater than 500:  : Total numbers are: 999 - 501 + 1 = 499

For the first position available options are:   and for the second and third position, we have 9 and 8 options.

Overall numbers with three distinct digits: 5 * 9 * 8 = 360

Overall numbers with all three digits same: 5

=> Three digit numbers with exactly one digit repeating: 499 - 360 - 5 = 134

Answer A

GmatPoint · Answer

Let a three-digit number abc.
Now, since only one digit is repeating, either a-b is the same, or a-c is the same, or b-c is the same.

Consider the case of a-c : For a the present limits, in a three digit number a-b-a, a can take 5 values(5-9) and b can take 9 values(0-9, except a).

Thus, total values = 9*5 = 45

Consider the case for a-b : For a the present limits, in a three digit number a-a-b, a can take 5 values(5-9) and b can take 9 values(0-9, except a).

Thus, total values = 9*5 = 45

Consider the case for b-c : For a the present limits, in a three digit number a-b-b, a can take 5 values(5-9) and b can take 9 values(0-9, except a).

Also, since 500 is excluded, total values = 9*5-1 = 45-1 = 44

Thus, the total number with exactly one digit repeating is 45+45+44 = 134

Thus, the correct option is A.

bumpbot · Answer

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Of the three digit numbers greater than 500, how many have exactly one

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GMAT Problem Solving (PS) Questions

Of the three digit numbers greater than 500, how many have exactly one

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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