If Whitney wrote the decimal representations for the first 300 positiv

Question

If Whitney wrote the decimal representations for the first 300 positive integer multiples of 5 and did not write any other numbers, how many times would she have written the digit 5 ?

A. 150
B. 185
C. 186
D. 200
E. 201

PS24851.01

chetan2u · Accepted Answer

Now, there are 300 multiples of 5..
(I)  Units digit...  All multiples of 5 will have either 5 or 0 as units digit, so half will have 5 in units digit..300/2=150
(II)  Tens digit..  The numbers are from 5*1 to 5*300 or 5 to 1500, so there are 15 hundreds and each hundred will have 5 as tens digit in two places.. A50 and A55
so 15*2=30
(III)  Hundreds digit...  5 will be in 100s place from A500 to A600, and as there will be 20 multiples of 5 in any given 100 numbers. 5 will be available 20 times from 500 to 595 and 1 time in 1500, so 20+1=21 times

Total =  150+30+21=201

E

ScottTargetTestPrep · Answer

The smallest multiple of 5 she wrote is 5, and the largest is 300 x 5 = 1,500. So Whitney could have written 5 as the units, tens or the hundreds digit.

Since every odd multiple of 5 has a units digit of 5 and every even multiple of 5 has a units digit of 0, we see that half of the 300 multiples, i.e., 150 of these multiples, will have a units digit of 5.

Now let’s investigate those that have a tens digit of 5. These numbers are 50, 55, 150, 155, …, 1,450, and 1455. We see that there are (1,450 - 50)/100 + 1 = 14 + 1 = 15 numbers that end with 0 and there should be another 15 that end with 5. So there are 30 numbers with a tens digit of 5.

Finally, let’s investigate those that have a hundreds digit of 5. These numbers are 500, 505, 510, … 595, and 1,500. We see that there are (595 - 500)/5 + 1 = 19 + 1 = 20 numbers in the 500’s that have a hundreds digit of 5, and including 1,500, there are 21 such numbers.

Therefore, Whitney must have written the digit five 150 + 30 + 21 = 201 times.

Answer: E

GMATinsight · Answer

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

https://www.youtube.com/watch?v=LrZEYFjQi5o

Joseph67890 · Answer

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

GMATinsight · Answer

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, 1 5 , 20, 2 5 , 30, 3 5 , 40, 4 5 ,  5 0,  55 , 60, 6 5 , 70, 7 5 , 80, 8 5 , 90, 9 5 , 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

GMATGuruNY · Answer

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

E

RK007 · Answer

How did u deal with the overlaps?

elavendan1 · Answer

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.

Babineaux · Answer

This is the way how I convinced myself.

Refer to the pic please.

Posted from my mobile device

dave13 · Answer

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 · Answer

"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

dave13 · Answer

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

BrentGMATPrepNow · Answer

"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

mcelroytutoring · Answer

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.

SVaidyaraman · Answer

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.

samarpan.g28 · Answer

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.

MalachiKeti · Answer

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

KarishmaB · Answer

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

Aboyhasnoname · Answer

Let's Try............. 
How many times we will write 5....... 
First number will be 5..Last will be 1500

Now: 
How many times 5 will appear in Units digit..... 
For all odd multiples........ So 150 times it will appear in Units Digit - I

How Many times 5 will appear in Ten's digit ... 
 2 Digit Number - 2 Times ----- Total Becomes 150 + 2 
 
3 Digit Number.... 
_ X _ X _ ....Units Place can be filled in 2 Ways....Ten's place in 1 Way (That is 5)..Hundreds Place in 9 Ways..(1-9)..So 9 x 2 = 18 times....Count becomes 150 + 18 +2 = 170

Four Digit Number 
_ X _ X _ X _ Units place can be filled in 2 ways, Tens place in 1 Way (That is 5)..Hundreds place in 5 Ways (0-4) and thousands place in 1 way.....2 x 1 x 5 x 1 = 10..Total becomes 180.....

How many times 5 will appear in hundred's digit.... 
3 Digit Number 
_ x _x _ ..Hundreds digit has 1 option (5) Ten's has 10 Options and Units has 2 Options.... 2 X 10 = 20 ..Total is 200

4 Digit Number 
So in 4 Digit Number the number with Hundreds place that will have 5 is only 1 Number that is 1500...Total becomes 201

If Whitney wrote the decimal representations for the first 300 positiv

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