How many three-digit positive integers have even number of even digits

Question

How many three-digit positive integers have even number of even digits?

A. 100
B. 125
C. 300
D. 325
E. 450

M37-73

CareerGeek · Accepted Answer

Even number of even digits 
—> either 0 even digits or 2 even digits should be there in the 3 digits

0 even digits —> all odd digits (OOO)
Number of ways = 5*5*5 = 125

2 even digits 
In this, 2 cases will come as 0 can’t be given to hundreds place

Case 1: when hundreds place is odd and other two are even (OEE)
—> Number of ways = 5*5*5 = 125

Case 2: when hundreds place is non zero even (2, 4, 6 or 8) I.e, (EOE or EEO)
—> Number of ways = (4*5*5)*2 = 200 [2 for EO or OE in tens and units places)

So, total = 125 + 125 + 200 = 450

IMO Option E

Bunuel · Answer

Official Solution:

How many three-digit positive integers have even number of even digits?

A.  100 
B.  125 
C.  300 
D.  325 
E.  450

We are looking for three-digit positive integers which have zero or two even digits. So, we are looking for three-digit positive integers which have all odd digits or two even digits and one odd digit.
 
The number of three-digit positive integers with all odd digits is  5*5*5=125  (5 odd options for each digit: 1, 3, 5, 7, or 9).
 
The number of three-digit positive integers with two even digits and one odd digit is a little bit trickier to get. We have to consider two cases:
 
If the first digit is even (EOE or EEO), then there will be  4*5*5+4*5*5=200  numbers (4 even options for the first digit: 2, 4, 6, or 8 because a three-digit number cannot start with 0).
 
If the first digit is NOT even (OEE), then there will be  5*5*5=125  numbers.
 
The total number of such numbers is therefore,  125+200+125=450

Answer: E

LevanKhukhunashvili · Answer

I counted 325

We need 2 digits even and 1 digit odd in 3 digit number

If the first digit is odd, we have 5 possibilities (1, 3, 5, 7, 9)
Then the rest two digits must be even, for each position we have 5 possibilities (0, 2, 4, 6, 8)
5*5*5=125 total numbers

If the first digit is even, we have got 4 possibilities (2, 4, 6, 8)
Then the arrangements for the rest two is even-odd or odd-even (2 possibilities)
For each even or odd digits, we have 5 possibilities (0, 2, 4, 6, 8 for even and 1, 3, 5, 7, 9 for odd)
4*5*5*2=200

125+200=325

IMO 
Ans:  D

Archit3110 · Answer

even = 0,2,4,6,8
odd= 1,3,5,7,9
we need 3 digit no which have even of even digits eg 100,122,223 so on
so if first digit is even we can have
4*5*5 = 100 *2 ways it can be arranged = 200
and if first digit is odd then ; 
5*5*5*2 ; 250; again 2 ways to arrange units and tens place
total 200+250 ; 450
IMO E

Shobhit7 · Answer

METHOD 1
Options:
1) EEO: 4*5*5= 100
2) EOE: 4*5*5= 100
3) OEE: 5*5*5= 125

Total: 325
Ans D

METHOD 2
Options: 
1) OOO: 5*5*5= 125
2) EEE: 4*5*5= 100
3) OOE: 5*5*5= 125
4) OEO: 5*5*5= 125
5) EOO: 4*5*5= 100

Sum: 575
Total Integers: 900
Remaining: 325

Ans D

Archit3110 · Answer

Shobhit7 
See OEE can be arranged in two ways for E eg 324,342.. 
125*2:250
Total 450 answer

r19 · Answer

need a details explanation of ques . my calculation was wrong then took a wild guess.

puneetb · Answer

I believe 324 and 342 has already been counted in 5*5*5
And it is absolutely correct with below:
1) EEO: 4*5*5= 100
2) EOE: 4*5*5= 100
3) OEE: 5*5*5= 125
4) OOO: 5*5*5=125

Total==325+125=450

menonrit · Answer

Hi According to " Case 2: when hundreds place is non zero even (2, 4, 6 or 8) I.e, (EOE or EEO)"

even OEE combo should be added so should you not be doing (4*5*5)*3

Shishou · Answer

The answer to this is 450.
We can have 0 even integers or 2 even integers.
If we have o even integers, then we have all 3 integers to be odd

Case I :ALL 3 ARE ODD:

since we can repeat the digits, we have the first odd number chosen in 5 ways, the second in 5 ways and the third in 5 ways as well,giving us:
5*5*5=125.

CASE II : 2 integers are even and 1 integer is odd

1. two of the even numbers are the same and 1 is odd
5C1 * 5C1* 3!/2!= 75 ways. However, we do not want the combinations that look like this:  0|0|odd number  and there are 5 ways such a combination could happen.
We also do not want those combinations that start like this  0|odd number|0  and there are 5 combinations that look like this.We will take out these ten combinations from 75 to get 65.

2. two different even integers and 1 odd number :
5C2 * 5C1 * 3!= 300 ways . However we do not want those combinations that look like this :  0|even|odd or 0|odd|even  and each of them can occur in 20 ways giving us 40 ways. We will take them out of 300 to get 260.

Hence total number of ways is 125+260+65=450 ways

Answer is E.

Kinshook · Answer

Asked: How many three-digit positive integers have even number of even digits?

Even digits = 2 or 0

0 Even (all odd) = 5*5*5 = 125

EEO = 4*5*5 = 100
EOE = 4*5*5 = 100
OEE = 5*5*5 = 125
EEO + EOE + OEE = 325

Such numbers (2 even & 0 even ) = 325 + 125 = 450

IMO E

jobu95 · Answer

I appreciate your solution, however a much simpler way to solve that problem exists:

Lets look at the odd/even ratio: 
The probability that the ratio is 3:0 is 1/4
The probability that the ratio is 2:1 is 1/4
The probability that the ratio is 1:2 is 1/4
The probability that the ratio is 0:3 is 1/4

Thus, (1/4 + 1/4) 900 is 450
Answer E

MHIKER · Answer

Three-digit even will be from 100-998
Even digit is divided by 2

So, the number of even digits =\frac{(998-100)}{2}+1  = 449+1= 450  is even number too. 
The answer is 450; E

Bunuel  I will appreciate if you help with my solution.

mysterymanrog · Answer

Hi,

While what you wrote is completely true, this is not what the question is asking. 
The question is asking how many 3 digit numbers have an even amount of even digits - not how many 3 digit numbers are divisible by 2. 
For example, the number 555 has an even number of even digits (0 is even), but this would not be included in your calculation.

The proper way to solve the question is the following:
We are looking for the number of 3 digit integers that have an even number of even digits. 
We can have two cases: either 2 even digits, or no even digits.
So cases are: OOO, OEE, EEO, EOE. Where E and O denote even/odd digit. 
the case of OOO is easiest - there are 5 odd digits, and therefore 5 choices for each slot 5^3=125. 
The case of OEE is the same as OOO. 5^3=125
The case of EEO has a small trick to it. Generally, you include 0 as an even digit, however because we have an even in the first digit, 0 cannot be included - otherwise it would be a 2 digit number, not a 3 digit (i.e 081). So the choices are now: 4*5*5 or simply 100. 
The case of EOE has the same trick as EEO. So another 100 possibilities
In summary we have 125*2+100*2=450 total cases.

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How many three-digit positive integers have even number of even digits

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