How many positive integers less than 500 begin with a 3, end with a 3,

Lets start with single digit numbers, we have only 1 number 3

Now 2 digit numbers we have of the form 3_ & _3
3_ are 10 numbers (30,31,32,33....39)
_3 are 9 numbers (13,23,33,.....93)
we get 33 in both sets, hence we remove 1 from total of 2 digit numbers,

Hence total 2 digit numbers is 18

Now 3 digit numbers, we have of the form 3 _ _ & _ _ 3
3 _ _ are 100 numbers (300,301,302,303,.....399)
_ _ 3 are 40 numbers (103,113,....203,213,....303,313,....413,423,...493)
we also have numbers of the form 3 _ 3 (303,313,...393) common to both sets, these are 10 such numbers.

Hence total 3 digit numbers is 100 + 40 -10 = 130

Hence the total numbers of the required form = 1 + 18 + 130 = 149.

Answer A.


Thanks,
GyM

Ps. You can also use combinatorics to count the numbers, for e.g. 3 _ _ can be 1*10C1*10C1 = 100 & _ _ 3 can be 4C1*10C1*1 = 40

The total number of 3 digit numbers that begin with 3 = 100.
The total number of numbers that end with 3 = From 0 to 100 - 10
From 100 to 200 - 10
From 200 to 300 - 10, Excluding all numbers ending with 3 in 300s as already counted.
From 400 to 500 - 10
The total number of numbers beginning with 3 but are 2 digit numbers = 9 (30, 31, 32, 34, 35, 36, 37, 38, 39) - Excluding 33 as was already counted in the ending with 3 section.
Total = 100 + 40 + 9 = 149.

We need to include one digit, two-digit, and three-digit numbers less than 500.

1-99
3 is the only one digit number
30-39 are the two digit numbers starting with 3. (Total of 10 numbers)
13,23,43,53,63,73,83,93 are the two-digit numbers ending with 3. (Total of 8 numbers)

\(100-199\)
113,123,133,143,153,163,173,183,and 193 are the two-digit numbers ending with 3 in this range

Similarly, both the ranges 200-299 and 400-499 have 10 numbers each which end with 3.

The range \(300-399\) has all the hundred numbers starting with 3(we needn't include those
numbers in this range which end with 3 because that would lead to double counting)

Therefore, there are a total of \(1 + 8 + 10*4 + 100 = 149\)(Option A) such numbers less than 500.

Yes, rishabhmail125 - Your observation is correct.

The numbers starting with 3 will include 33 in the list. The total numbers starting with 3 is 111.
Also, 3 is part of both the lists. The total numbers which are part of both lists are 12 in number.

Therefore, the total numbers(starting with 3, ending with 3, or both) less than 500 are 149(111+50-12)

Rishabh, you are absolutely correct..
There was a typo in the post, edited now.
There will be 111 positive integers starting with 3 and 50 ending with 3 but 12 common to both so 111+50-12=149

There are 499 integers from 1 to 499 inclusive.

Let's determine how many of those 499 integers DO NOT meet the condition of beginning with a 3, ending with a 3, or both

1-digit integers that DO NOT begin with a 3, end with a 3, or both
1, 2, 4, 5, 6, 7, 8, and 9
TOTAL = 8

2-digit integers that DO NOT begin with a 3, end with a 3, or both
In the tens position, we can have 1, 2, 4, 5, 6, 7, 8, or 9 (8 possibilities)
In the units position, we can have 0, 1, 2, 4, 5, 6, 7, 8, or 9 (9 possibilities)
TOTAL number of 2-digit integers that DO NOT begin with a 3, end with a 3, or both = (8)(9) = 72

3-digit integers that DO NOT begin with a 3, end with a 3, or both
In the hundreds position, we can have 1, 2, or 4 (3 possibilities)
In the tens position, we can have 0, 1, 2, 3, 4, 5, 6, 7, 8, or 9 (10 possibilities)
In the units position, we can have 0, 1, 2, 4, 5, 6, 7, 8, or 9 (9 possibilities)
TOTAL number of 3-digit integers that DO NOT begin with a 3, end with a 3, or both = (3)(10)(9) = 270

So, the TOTAL number of integers that DO NOT meet the given condition = 8 + 72 + 270 = 350

TOTAL number of integers that MEET the given condition
So, the = 499 - 350 = 149

Answer: A

Cheers,
Brent

Sharing my approach to this question using the slot method:

In all there are \(10\) digits from \(0, 1, 2, 3, ... , 9\) inclusive

Single Digit Number \(= 1\)
Only \(3\)

Double Digit Numbers \( = 18\)
a. Starts and ends with 3 \(=\) Only \(33\)
b. Starts with 3 but ends with a number other than 3 \(=\) 1 9 \(= 9\)
c. Starts with a number other than 3 but ends with 3 \(=\) 8 1 \(= 8\) (The first digit cannot be 0 and 3, hence 8)

Three Digit Numbers \(= 130\)
a. Starts and ends with 3 \(=\) 1 10 1
b. Starts with 3 but ends with a number other than 3 \(=\) 1 10 9 \(= 90\)
c. Starts with a number other than 3 but ends with 3 \(=\) 3 10 1 \(= 30\) (Only 1, 2 and 4 can take the first place as the number has to be between 100 and 500)

Combining the results = 1 + 18 + 130 = 149 (Ans. A)

Another fun one. Key is to avoid over counting and try to keep straight the numbers that repeat the digit 3 in the front and back (33 and 303) as over-counting either of these Numbers will land you with the wrong answer.

(1st) numbers that begin with a 3 digit

(A)One digit numbers - only 3

1 number

(B) Two digit numbers that begin with 3

Tens place: 1 available option as 3 ——-1
And
Units place: every digit except 3 ——— 9

1 * 9 = 9 Numbers

(C) Three Digit Numbers that begin with 3:

Hundreds Place: 1 available option, locked in 3 ———- 1

And

Tens Place: can take any of the 10 digits ——- 10

And

Units Place: every digit except 3 ———9

1 * 10 * 9 = 90 Numbers

So far we have: 1 + 9 + 90 = 100 Numbers.



(2nd) Numbers that End with a 3 digit (except the single digit number 3)

From 0 to 100: (note: we did not count 33 above, so we will count it here)

13, 23, 33, 43, 53, 63, 73, 83, 93 ——> 9 Numbers

103, 113, 123, ........ 193 ———-> 10 Numbers

203 .... 293 ———-> 10 Numbers

303 .....393 (note: we did not include digit 3 above when we counted numbers that began with 3) ———> 10 Numbers

403 ..... 493 ———-> 10 Numbers

9 + 10 + 10 + 10 + 10 = 49


Finally, there are - 100 + 49 =

(A) 149 Numbers

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The number with hundreds digit as 3. From 300 to 399 = 100 -- (1)
The number with units digit as 3. 10 (in each hundred) x 5 = 50--(2)
Less one number taken twice i.e. 333. considered in both (1) and (2)--- (3)

The correct answer in Option A. (1) + (2) - (3) = 100 + 50 -1 = 149.

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