If Whitney wrote the decimal representations for the first 300 positiv

Please check the video for the step-by-step solution.

Positive 300 multiples of 5's are between 5-1500
No of 5's as units digits = 150 (5,15,25........)
No of 5's as tens digit = 30 (50,55,150,155.......)
No of 5's as hundreds digit = 21 (500,505,510,515....and 1500)

Total = 201

Method-2

300 multiples of 5 will be spread among first 1500 numbers (from 1 to 1500) because 300th multiple of 5= 300*5=1500

Digits used in multiples of 5 from 1 to 100 = 12
{5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90, 95, 100}

Every 100 numbers will use digit 5 12 times at unit and tens place

so total usage of digit 5 in 15 centuries of numbers = 12*15 = 180

At hundred's place 5 will be used 20 times in 20 multiples of 5 from 500 to 595

One more 5 will be used at hundred's place in 1500

Total digit 5 use = 180+20+1 = 201

ANswer: Option E

Units place:
The units digits alternate between 0 and 5.
Implication:
Among the 300 values, the probability of getting a 5 in the units place is \(\frac{1}{2}\).
\(\frac{1}{2} * 300 = 150\)

To make it easier to count the remaining appearances of 5, represent each number up to 1495 as a 4-digit value, starting with 0000:
0000, 0005, 0010, 0015...0980, 0985, 0990, 0995...1480, 1485, 1490, 1495

Tens place:
In the blue list above, each digit 0-9 will appear in the tens place the same number of times.
Implication:
Among the 300 values, the probability of getting a 5 in the tens place is \(\frac{1}{10}\).
\(\frac{1}{10} * 300 = 30\)

Hundreds place up to 0995:
Number of multiples of 5 between 0000 and 0995, inclusive:
\(\frac{biggest-smallest}{increment }+ 1 = \frac{0995-0000}{5} + 1 = 199 + 1 = 200\)
Up to 0995 in the blue list above, each digit 0-9 will appear in the hundreds place the same number of times.
Implication:
Among these 200 values, the probability of getting a 5 in the hundreds place is \(\frac{1}{10}\).
\(\frac{1}{10} * 200 = 20\)

Hundreds place between 1000 and 1500, inclusive:
Only 1500 will yield a 5 in the hundreds place.
Number of cases = 1

Total number of 5's = 150+30+20+1 = 201

"decimal representation" is just one way to say that the numbers are NOT expressed as fractions.
As you might imagine, there are many different ways to express the first 300 multiples of 5.
For example, we could express the first multiple as 5, in which case we write the digit 5 one time.
However, we could also express the first multiple as 25/5, in which case we've written the digit 5 two times.
Or we could also express the first multiple as 555/111, in which case we've written the digit 5 three times.
etc.

As for solutions, I don't think I can do any better then the solution that chetan2u provided here: https://gmatclub.com/forum/if-whitney-w ... l#p2392437

How did u deal with the overlaps?

There will not be overlaps. It may sound absurd, but here we consider the 5's as units/tens/hundreds. i.e, unit digit appearance, tens digit appearance and hundreds digit appearance.

For example, consider 555. Our idea here is to count the number of 5's written. So, the unit digit will be counted during Step 1, tens digit during Step 2 and hundreds digit during Step 3.

This is the way how I convinced myself.

Refer to the pic please.

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hi BrentGMATPrepNow

i m bit confused by term " decimal representation of integer" is 55 = 5*10+5 just like 125 = 1*100+2*10+1*5

can you please provide your solution

BrentGMATPrepNow thanks but now i got totally confused cause my imagination is ulimited so i better stick to what is implied in the question

what does this mean ? If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5

for example 55 = 5*10 +5 (here we write 5 two times) 15 = 10*1+5*1 (here 5 is written one time) is it correct how i should approach the problem?

is it how how should understand the highlighted part
chetan2u

"decimal representation" just means expressing the number in the form XXXX.XXX
For example, all of the following are decimal representations of the number 5: 5, 5.0, 5.00, 5.000, 5.0000, ....etc

So, this definition rules out representing 5 as 5 + 15 - 10, etc

Let's further explore the awkward term "decimal representation" in this question, whose wording implies that the numbers in question should be written in some sort of fractional / decimal form, when in fact they are simply regular integers.

You're better off ignoring the term "decimal representation" in this particular case. This question is using the broader definition of the word "decimal," which is the adjective part of speech: "relating to or denoting a system of numbers and arithmetic based on the number ten, tenth parts, and powers of ten." In other words: despite the deceptive wording, a "decimal representation" can in fact be any number, even integers.

Please note that the word "decimal" almost always indicates a real decimal on today's GMAT. This question is a rare exception — possibly because of the advanced age of the question. Unfortunately, many of the so-called "GMAT Advanced Questions" (the book was published in 2019) are in fact quite old — in fact, some go all the way back to 1990s editions of the Official Guide, way back when it was written by ETS, not GMAC.

There are 150 multiples of 5 that end with 5. Now to make things simple, find the ratio of number of 5's in units place and number of 5's in tens place . In the range 5 to 95, the ratio is 5:1. So over the complete range, there are 150/5 = 30 fives in tens place. Similarly the ratio of 5's in hundreds place to tens and units place in the range 500 to 595 is 1:1 and 1:1 resp. So there are 20 5s in hundreds place. Then we have the 5 in 1500, for a total of 150+30+20+1= 201.

­First 300 positive integer multiple of 5 means 5 to 1500. We need to find the number of times 5 as a digit will repeat in the set.
1. From 1 to 100: 5,10,15,20,25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100 - 5 repeats 12 times.
2. Similarly for the ranges 101 to 200, 201 to 300, 301 to 400 - 5 repeats 12 times.
3. For 401 to 500, 5 will be there 13 times because we need to consider the '5' in 500 as well.
4. 501 to 600 will have 5 31 times.
5. Again, the ranges 601 to 700,....,1401 to 1500 will have 12 5s.
6. Finally, 1500 will have only one 5.
Therefore, 12*13 + 13 + 31 + 1 = 201. Option (E) is correct.

I didnt feel comfortable with other approaches so used arrangements principle instead

1. 5 in units digit

Case 1: 5 in units digit for 1,2 and 3 digits multiples
__ ___ __
10* 10*1 = 100
OR
Case 2: 5 in units digit for 4 digit multiple upto 1499 (1500 anyway doesnt have 5 in units digit)
__ __ __ __
1*5*10*1 = 150
Total times 5 is in units digit is = 150

2. 5 in tens digit

Case 1: 5 in tens digit for 1,2 and 3 digits multiples
__ ___ __
10* 1*2 = 20
OR
Case 2: 5 in tens digit for 4 digit multiple upto 1499 (1500 anyway doesnt have 5 in tens digit)
__ __ __ __
1*5*1*2 = 10
Total times 5 is in tens digit is = 20+10 = 30

3. 5 in hundreds digit
Case 1: Three digit multiples of 5 starting with 5
__ __ __
1*10*2 = 20
OR

1500
Total times 5 is in Hundreds digit is = 20+1 = 21

Adding up all 6 cases will give us final answer 150+30+21 = 201

First 300 positive multiples of 5 will range from 5 to 1500.

We know that in first 100 consecutive positive numbers, each digit other than 0 appears 20 times: 10 times in units digit and 10 times in tens digit.
We are writing only the multiple of 5 so all 10 times that it appears in the units digit, we are writing those out.
But when it is appearing in the tens digit, we are only writing it out when the units digit is 0 or 5. Hence we are writing 5 in tens place only twice.
So there will be 12 appearances of 5 in units and tens digit in every 100 numbers.

Then how many times will we write 5 in units and tens digits? 12 * 15 = 180 times

But 5 can appear in the hundreds place too in the 500 - 599 range. The 20 multiples of 5 that appear in these 100 integers will be written and hence we will get another 20 5s which will appear in the hundreds place. So total becomes 180 + 20 = 200

But we haven't accounted for the 4 digit number 1500 in which 5 will appear in the hundreds place too. So finally the total becomes 200 + 1 = 201

Answer (E)

Here is a discussion on some other such questions: https://youtu.be/_C-kTA44OxE