Integers

imo E

basically the tens digit = 1
units digit = 3

i.e. the number should be - - - - 13

every one hundred digits have only one number which has tens digit 1 and units digit 3

therefore

458600 - 324700 = 133900

133900/100 = 1339 (as every one hundred digits have only one number which has tens digit 1 and units digit 3)

rohangupta83 wrote:imo E

every one hundred digits have only one number which has tens digit 1 and units digit 3

CAn you plz. explain this step..........


Thanks

You want to basically count how many numbers have 13 at the end, between 324,700 and 458,600.

- Let's ignore 324,700-324,999 and 458,001-458,600 to start, and just say: "try and find how many 13's there are between 325,000 and 458,000."

- You know that each thousand number has 10 of them (ie 325,013; 325,113; 325,213....325,913 etc).
- You also know that there are 133 "thousands" between 325,000 and 458,000
- so if each of these "thousands" has 10 "13s", then we have a total of 1,330 "13's".
- but remember, we have to add the few we ignored at first: so there are 3 "13's going from 324,700-324,999 [ie 324,713; 324,813; 324,913] and 6 of them going from 458,000 to 458,600, for a total of 9.

So, 1330 + 9 = 1339.

Hope that helps.

anayeri wrote:You want to basically count how many numbers have 13 at the end, between 324,700 and 458,600.

- Let's ignore 324,700-324,999 and 458,001-458,600 to start, and just say: "try and find how many 13's there are between 325,000 and 458,000."

- You know that each thousand number has 10 of them (ie 325,013; 325,113; 325,213....325,913 etc).
- You also know that there are 133 "thousands" between 325,000 and 458,000
- so if each of these "thousands" has 10 "13s", then we have a total of 1,330 "13's".
- but remember, we have to add the few we ignored at first: so there are 3 "13's going from 324,700-324,999 [ie 324,713; 324,813; 324,913] and 6 of them going from 458,000 to 458,600, for a total of 9.

So, 1330 + 9 = 1339.

Hope that helps.

Very clear.......
Thanks