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

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Joined: 12 Sep 2015
Posts: 4000
Re: How many integers between 324,700 and 458,600 have tens  [#permalink]

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27 Feb 2018, 13:02
Top Contributor
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

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

Cheers,
Brent
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Re: How many integers between 324,700 and 458,600 have tens  [#permalink]

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12 Sep 2019, 01:15
144144 wrote:
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.

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
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Re: How many integers between 324,700 and 458,600 have tens  [#permalink]

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27 Sep 2019, 03:25
kanikab wrote:
144144 wrote:
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.

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

Did you count in 333313 ?

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.
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Karishma
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Re: How many integers between 324,700 and 458,600 have tens   [#permalink] 27 Sep 2019, 03:25

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