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

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)

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.

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!

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.

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

Thanks in advance

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.

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: 13 (1 integer)
Integers from 101 to 200: 113 (1 integer)
Integers from 201 to 300: 213 (1 integer)
Integers from 301 to 400: 313 (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

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

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.

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?

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.

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

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