Problem Solving

fiza gupta wrote:How many integers between 324,700 and 458,600 have a 2 in the tens digit and 1 in the units digit?

(A) 1,339
(B) 1.352
(C) 1,353
(D) 10,030
(E) 10,300

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 2 in the tens digit and 1 in the units digit.
Integers from 1 to 100: 21 - 1 integer
Integers from 101 to 200: 121 - 1 integer
Integers from 201 to 300: 221 - 1 integer
Integers from 301 to 400: 321 - 1 integer
.
.
.
As we can see, for every 100 integers, there's 1 integer with a 2 in the tens digit and 1 in the units digit.
In other words, 1/100 of all integers have a 2 in the tens digit and 1 in the units digit.

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

Cheers,
Brent $$$$

santhosh_katkurwar wrote:Hi Brent,

Is it possible to find through method after finding the number of integers?

Thanks,
Santosh

Hi Santosh,

I'm not sure what you're asking.
Once I determined that there are 133901 integers from 324,700 to 458,600, I used the fact that 1/100 of those integers end in 21 to arrive at the correct answer.

Cheers,
Brent

fiza gupta wrote:How many integers between 324,700 and 458,600 have a 2 in the tens digit and 1 in the units digit?

(A) 1,339
(B) 1.352
(C) 1,353
(D) 10,030
(E) 10,300

The first few integers that have this property are: 324,721, 324,821, 324,921, .... As we can see, this is an arithmetic sequence with common difference of 100. Since the last integer that has this property is 458,521, the number of integers that have this property is:

(458,521 - 324,721)/100 + 1 = 133,800/100 + 1 = 1,338 + 1 = 1,339

Answer: A