How many times digit 6 is used while writing numbers from 100 to 1000?

Question

How many times digit 6 is used while writing numbers from 100 to 1000?
(A) 648
(B) 280
(C) 252
(D) 225
(E) 26

Source:  https://www.GMATinsight.com

MathRevolution · Answer

Forget conventional ways of solving math questions. In PS, IVY approach is the easiest and quickest way to find the answer.

How many times digit 6 is used while writing numbers from 100 to 1000?
(A) 648
(B) 300
(C) 252
(D) 225
(E) 26

The digit 6 is used once in 100~109 (only when 106).
Similarly the digit 6 is used once in 110~119(only when 116),
 .... 
 in 150~159(only when 156),
 in 170~179(only when 176),
 in 190~199(only when 196). 
On the other hand in 160~169 the digit 6 is used at unit digit(166) and at tenth digit(16□). 
So the digit 6 is used 11 times(=1 (unit digit)+10 (tenth digit)) in 160~169.

That means the digit 6 is used 20 in 100~199.

Similarly in 200~299 : 20 times
 300~399 : 20 times
 ....
 500~599 : 20 times
 700~799 : 20 times
 ....
 800~899 : 20 times
 900~999 : 20 times
On the other hand in 600~699 we should consider hundredth digit so 6 is 100 times more used. 
That means the digit 6 is used 120 in 600~699.

So total number of digit 6 is 20*10 + 100= 300.

The answer is (B).

hdwnkr · Answer

Although combinations method can be used, I used a different approach that helped me solve this in less than a minute. A close look at the options tell us that A is too much. E is too Less. Now,

There are 100 numbers which begin with 600
Next, In every 10 numbers such as 100 to 110, 110 to 120, 120 to 130 6 comes at least once. Number of such intervals = End limit - First no. / interval. Our range of numbers is 100 - 1000
1000 - 100 = 900/10 = 90
Number of 10s interval in this is 90. So 90 '6s'

So far we have calculated 190.

Now, 60 - 69 is the tens interval that has 11 '6s' (becasue 66 has two '6s), 1 of which has been counted in the previous step. And there are 9 such intervals (160 to 169, 260 to 269, 360 to 369, ...... ,960 to 969) This gives us another 90 '6s'

The total now comes to 280. The nearest to which is 300. Hence B.

(A) 648
(B) 300
(C) 252
(D) 225
(E) 26

Praveengeol · Answer

Silly query may be -

I am getting 280, may be i am making some mistake. Please correct me and let me know where i am making mistake.

I am taking 100 jintervals each from 100 to 199 and it gives me 9 intervals and not 10 as you have taken. So, as we have 20 instances of using digit 6, we can say 20*9 = 180 and 100 more shoul dbe added for the multi use of digit 6 between 600 to 699.

So, i get answer 280. On test day, i will mark the nearest as 300 but still will be better to get a right answer while practice.

I am sure i am making mistake somewhere, please correct.

Thanks

hwgmat2015 · Answer

I did it this way Number of 6's from 100 -200 => 1 * 1 * 9 = 9 ( 161,169, except 166.) 1 * 9 * 1 = 9 ( 116,126 except 166) 166 => 2 ( 2 sixes, 166) Therefore 9 + 9 + 2 =20 Now 100 - 999 ( Except 600 - 699) = 20 * 9 =180 Number of 6's from 600 - 700 601 = 1 * 9 * 9 = 81 661 = 1 * 1 * 9 = 9 * 2 ( two 6's)= 18 616 = 1 * 9 * 1 = 9 * 2 ( two 6's) =18 666 = 03 ----------------------------------------------- Total = 120 total number of 6's = 180 + 120 = 300 IMO : B Kudos if this solution helped you !!

sowragu · Answer

Possibility of 6 appearing in a three digit number 
It can appear only once in three digit number   or two times in a three digit number   or three times  .

Hence we can use the combination.

If six appearing only once in the 100th place = 1*9*9 = 81. Since we have 10th and unit place we can multiply by 3. => 3 * 81 = 243
Similarly for the appearance of two times = 1*1*9 = 9 times two 6's appearing means we have 18 6's. Since we can have three possibilities as shown above ==> 3*18 = 54
three times 6 can appear once = 1*1*1 = 1 which has 3 6's

Hence 243+54+3 = 300.

wihi · Answer

Has anybody ever tried to check this task?

The list of answers doesn't contain the correct one: 280.

We can just calculate the number of 6's in hundreds, tens and units digits. By using this approach, we are not going to miss or double count any occurrence of 6.
Let's count:
6** - 100 (600-699)
*6* - 90 (9 options for hundreds and 10 options for units)
**6 - 90 (9 options for hundreds and 10 options for tens)

The total is  280 , not 300.

That's it. There are no more occurrences of 6 on the interval from 100 to 1000.

AR15J · Answer

Hi  Bunuel ,

I also got 280 as the digits need to be count between 100 and 1000. If the interval were 0 to 1000, then the answer would be 300 , but for the given interval, the answer should be 280. Please confirm.

leanhdung · Answer

Instead of counting from 100 to 1000, we should count from 100 to 999 with no change in the result (the number 1000 does not contain any 6).

Each number from 100 to 999 has 3 digits, and there are 999 - 100 +1 = numbers from 100 to 999. Hence, there are total 1000*3 = 3000 digits from 100 to 999. We also know that each number in {0,1,2,3,4,5,6,7,8,9} has an equal chance of 1/10 to appear in the total 3000 digits, so number 6 appears in 3000*(1/10) = 300 times!

Done!

Please kudos me if you think my explanation is good

wihi · Answer

1) No, 999 - 100 +1 is 900, not 1000.
and
2) No, the chance is not 1/10. There is no digit 0 in hundreds, because the numbers are starting from 100.

leanhdung · Answer

Yes, I made a mistake! Here is my new solution:

From 100 to 999, there are 999 - 100 + 1 = 900 numbers. Hence, we have 900*3 = 2700 digits!

We can observe that there is no digit 0 in hundreds, but each number in {1,2,3,4,5,6,7,8,9} (excluding number 0)still has an equal chance to appear in the total 2700 digits.

First, we count the number of 0 = 9*(2+9*2) = 180, let x is the number of 6 that appears in the total 2700 digits --> 9*x+180 = 2700 --> x =180

---> D

Done!

louisbharnabas · Answer

All,

The answer is 280 (for 100 to 1000).

Whereas the answer is 300 (for 0 to 1000)

Method:

Ignore 1000 and take only values from 100 to 999.

1st digit can be 1 to 9 (9 possibilities). Note: 0 is not included. If 0 is included it considers 0 to 99 as well. 
2nd digit can be 0 to 9 (10 possibilities)
3rd digit can be 0 to 9 (10 possibilities)

Case 1: 1st digit is 6:

1st digit is 6 (1 possibilities). 
2nd digit can be 0 to 9 (10 possibilities)
3rd digit can be 0 to 9 (10 possibilities)

=> 1 x 10 x 10 = 100

Case 2: 2nd digit is 6:

1st digit can be 1 to 9 (9 possibilities). Note: 0 is not included. If 0 is included it considers 0 to 99 as well. 
2nd digit is 6 (1 possibilities)
3rd digit can be 0 to 9 (10 possibilities)

=> 9 x 1 x 10 = 90

Case 3: 3rd digit is 6:

1st digit can be 1 to 9 (9 possibilities). Note: 0 is not included. If 0 is included it considers 0 to 99 as well. 
2nd digit can be 0 to 9 (10 possibilities)
3rd digit is 6 (1 possibilities)

=> 9 x 10 x 1 = 90

Add all cases:

100 + 90 + 90 = 280

Regards.

AR15J · Answer

Hi  VeritasPrepKarishma

Please confirm that the correct answer is 280. Thanks a lot .

Bunuel · Answer

Yes, the answer should be 280.  GMATinsight  please revise the options/question. Thank you.

GMATinsight · Answer

Fixed the option. Thankyou...

Archit3110 · Answer

total digits from 100 to 999 ; 900
target how many times digit 6 is used in 100 to 1000 ; 
number of times digit 6 comes in a series of 100 to 199 ; 1*9*9 ; 81 ; 100-81 ; 19 + ( 1*1*1 i.e 166) ; 20
20*9 ; 180 terms of 6 at units & tens places
6 digit at hundereds place 600 to 699; 100 
total times 6 digit comes from 100 to 1000 ; 180+100
280 option B

How many times digit 6 is used while writing numbers from 100 to 1000?

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How many times digit 6 is used while writing numbers from 100 to 1000?

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