How many integers between 324,700 and 458,600 have tens

Question

How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?

A. 10,300
B. 10,030
C. 1,353
D. 1,352
E. 1,339

Bunuel · Accepted Answer

There is one number in hundred with 1 in the tens digit and 3 in the units digit:  13 , 1 13 , 2 13 , 3 13 , ...

The difference between 324,700 and 458,600 is 458,600-324,700=133,900 - one number per each hundred gives 133,900/100=1,339 numbers.

Answer: E.

EvaJager · Answer

The numbers must be of the form ABCD13, where ABCD is any four-digit number between 3247 and 4585 inclusive. 
ABCD cannot be 4586, as 458,613>458,600.

Therefore, the total number is 4585 - 3247 + 1 = 1339.

Answer E

144144 · Answer

is there another way to solve this?

I once saw someone solving this kind of questions by multiplying the amount of different number of digits u can put in each place (in each different digit)

KarishmaB · Answer

You can but you would have to take multiple cases and that would be really cumbersome. The left most digit can be either a 3 or a 4 but then next digit depends on what the leftmost digit is. If left most digit is 3, next digit can be anything from 2 to 9. If the leftmost digit is 4, the next digit can be from 0 to 5. Similarly other digits too.
So preferably, focus on the approach given by Bunuel.

Perhaps · Answer

m clear with the statement that there are 133,899 integers between the two numbers. but how to calculate hundreds of numbers in them..... There are 1338+1 (to account for the number 133,821-------cud not understand this particular step.... If anyone can explain me this step in little detail then plz explain coz i dont hv any idea

Thanks in advance!

mau5 · Answer

When we have to find the numbers between A and B(where both A and B are included) = (B-A)/1 +1.

When we have to find the numbers between A and B(where both A and B are excluded) = (B-A)/1 -1.

When we have to find the numbers between A and B(where either A or B is included) = (B-A)/1 .

For a number to keep having a 2 in the tens place and 1 in the units place, we have to keep adding 100 to that number. Thus, starting from 324,721, keep adding 100 to get the same units and digits place. Now to find the hundreds, all we have to do is find the number of hundreds between 458,521 and 324,721, where both are included : (458521-324721)/100+1 = 1338+1 = 1339.

GGMAT760 · Answer

Hi
Why you divided by 100..Please explain..I read all below comments but still can not understand the concept.

Thanks in advance

Bunuel · Answer

In each hundred there is only one number with 1 in the tens digit and 3 in the units digit. How many hundreds are in 133,900? 133,900/100=1,339. One number per hundred gives total of 1,339*1 numbers.

BrentGMATPrepNow · Answer

A nice rule says:   the number of integers from x to y inclusive equals y - x + 1  
So, the number of integers from 324,700 to 458,600 = 458,600 - 324,700 + 1 =   133901  
ASIDE: We can safely round this number   133900   (you'll see why shortly)

Now let's look at how frequently we have a  1  in the tens digit and  3  in the units digit.
Integers from 1 to 100:  1  3  (1 integer) 
Integers from 101 to 200: 1 1  3  (1 integer) 
Integers from 201 to 300: 2 1  3  (1 integer)
Integers from 301 to 400: 3 1  3  (1 integer) 
.
.
.
As we can see, for every 100 integers, there's 1 integer with a  1  in the tens digit and  3  in the units digit.
In other words, 1/100 of all integers have a  1  in the tens digit and  3  in the units digit.

So, among the   133900   integers in question 1/100 of them meet the given condition. 
(1/100)(  133900  ) = 1339

Answer: E

Cheers,
Brent

kanikab · Answer

VeritasKarishma  
Hi, I tried this method but got an incorrect answer. I saw someone else also did this but I didnt get the solution. Can you please help?

Here is what i did.

For numbers starting with the digit 3, it will be 1(because only 3 can come here)*8*(because any number above 2 can come here)*6(any number above 4 can come here)*3(any number above 7 can come here)*1*1.

Similarly for numbers starting with 4 using the same logic above - 1*6*9*7*1*1

The answer should be adding the 2 combinations and subtracting 1 from it which is 521

KarishmaB · Answer

Let me ask you this:

Did you count in 333313 ?
What about 334613 ?

When my second digit from the left is 3, my third and fourth digits can be anything, not necessarily greater than 4 and 7 respectively. 
It is far too cumbersome.

Hoozan · Answer

Bunuel   Vinit800HBS  We have 324700 and 458600 so we know that our first number that satisfies the given condition will be 324713 and our last number will be 458513

But now if we consider the approach that you mentioned then we get:

458600 - 324700 --> 133800

So, 133800/100 = 1338

Could you explain why do we get a different result? I agree that we have taken two different start and end points in our set BUT shouldn't that NOT affect our solution?

Vinit800HBS · Answer

Hi Hoozan,

You made a very silly calculation mistake.

458600 - 324700 --> 133900

It will indeed affect the answer. 
Take for example this:
If you want to calculate _13 in the range 200-800, then numbers are 213, 313, 413, 513, 613, 713 ---> 6 cases.

So, ideally, you should do 713-213/100 +  1(to include the extreme terms)  = 6
But if you do 813-213/6 (which, I am assuming, you meant by 800-200, probably you thought that 13 will anyway cancel out), the problem is you are accounting for scenario (813) which is beyond the range.

Please do let me know if I have misunderstood your question.

paulrahul · Answer

last 2 digits are fixed at 13 => This occurs once for every 100 numbers, ie 113, 213, 313....

Form AP with first term as 324713 and last term as 458513 and common difference as 100.
nth term of AP = first term + (n-1)*common diff
=> 458513 = 324713 + 100*(n-1)
=> 133800 = 100 * (n-1)
=> n = 1339

E is the correct answer

Mantrix · Answer

Can Solve by AP

324713=a=first term, 324813=second term, so difference=d=100
458513 last term

now 
a+(n-1)d = Last term
324713+(n-1)100 = 458513
(n-1)100 = 133800
(n-1) = 1338
n= 1339

AnkurGMAT20 · Answer

Bunuel  do you have more such problems to practice?

Bunuel · Answer

Similar questions to practice: https://gmatclub.com/forum/how-many-int ... 96934.html https://gmatclub.com/forum/how-many-int ... 85397.html Hope it helps.

MalachiKeti · Answer

I know noone wants this solution, but if there is any weirdo who is interested the PNC answer to this here it is. Arrived in 10 minutes cause kept missing cases.

We have 3 cases for 6 digit numbers starting with 3 and ending with 13 and greater than 324700
Case 1: 3 2 4 _ 13
Case 2: 3 2 _ _ 13
Case 3: 3 _ _ _ 13

Case 1: Only 3 digits can be in hundreds place as we have to think of numbers greater than 324700. Those 3 digits are 7,8,9. Why 7? Because 324713>324700. Case 1 = 3 possibilities

Case 2: Only 5 digits can be in thousands place (5,6,7,8,9). 4 we have already accounted for and Digits (0,1,2,3) will be less. For Hundreds place we can have all 10 digits. Case 2 = 5*10 = 50 Possibilities

Case 3: All digits in ten thousands place that are not 0,1,2 so 7 cases and we can have 10 digits for thousands and 10 digits for hundreds hence 7*10*10 = 700 cases

Total of 3 cases = 3+50+700 = 753 --------(I)

We have 3 cases for 6 digit numbers starting with 4 and ending with 13 and lesser than 458600
Case 1: 4 5 8 _ 13
Case 2: 4 5 _ _ 13
Case 3: 4 _ _ _ 13

Case 1: Only 6 digits can be in hundreds place as we have to think of numbers lesser than 458600. Those 6 digits are 0,1,2,3,4,5. Why not 6? Because 458613>458600 which violates our condition. Case 1 = 6 possibilities

Case 2: Only 8 digits can be in thousands place (0 to 7). 8 we have already accounted for and Digits (9) will be more. For Hundreds place we can have all 10 digits. Case 2 = 8*10 = 80 Possibilities

Case 3: All digits in ten thousands place that are not 5 to 9 hence are (0,1,2,3,4) so 5 cases (5 we have already tackled separately) and we can have 10 digits for thousands and 10 digits for hundreds hence 5*10*10 = 500 cases

Total Cases: 6+80+500 = 586 ------- (ii)

Total cases = 586+ 753 = 1339

Kindly dont solve by this method:)

MalachiKeti · Answer

Hey  Bunuel  when we are talking about between we are talking about the exclusivity case rigt..where both nos are not included. While it wont matter in this case but the no of integers between the two digits should have been 133899 right?

How many integers between 324,700 and 458,600 have tens

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How many integers between 324,700 and 458,600 have tens

Prep Toolkit

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