Around the World in 80 Questions (Day 9): A number is selected at

The last three digits of a number determine whether it is divisible by 8 or not.
For a four-digit number with all even digits, there are 5 choices (0, 2, 4, 6, 8) for each of the first two digits, but only 4 choices for the third and fourth digits (2, 4, 6, 8) because 0 is not allowed.
Therefore total number of such four-digit numbers is 5 * 5 * 4 * 4 = 400.
For a number to be divisible by 8, the last three digits must be divisible by 8. The possible combinations are 000, 008, 020, 024, ..., 888, which gives us 50 combinations.
Hence the probability that a randomly selected four-digit number with all even digits is divisible by 8 is 50/400 = 1/8.
Hence A

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 8/25
(E) 2/5

Four digit positive number can be from 2,4,6,8 so it would be 1 out 4.

ways to have 4 digit number
0,2,4,6,8
repetition not allowed ; 4*4*3*2 ; 96
repetition is allowed 4*5*5*5 ; 500
but for a number to be selected and be divisible by 8 has to be among 0,2,4,6,8
1/5
and since we have two ways i.e. repeating and non repeating
2,4,6,8 ;1/4
1,3,8,8 ; 2/4
1,8,8,8 ; 3/4
8,8,8,8, ; 4/4
1/4 * 2/4*3/4 * 4/4 ; 6/64 ; 3/32 * 2c1 ; 3/16
option B

The 4 digit number has only even digits.

So the total numbers can be:
_ _ _ _
The 1st digit can be any of 2, 4, 6, 8.
All other digits can be any of 0, 2, 4, 6, 8.

The total digits that can be formed are 4 * 5 * 5 * 5 = 500.

Now, although for number to be divisible by 8, last 3 digits should be divisible by 8, but the same can also be identified by seeing the cyclic nature. 2 in 5 number block is divisible by 8. So the probability of the number being divisible by 8 is 2/5.

Number of 4-digit integers that have all their digits as even numbers = 4 * 5 * 5 * 5

For a number to be divisible by 8, its last three digits should be divisible by 8.
So lets calculate three digit numbers consisting of only even numbers divisible by 8. Include the numbers starting with 0 such as 024, 048 etc.

Number of such possibilities = 8 * 5 * 4

Probability = \(8/25\)

Ans D

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

4 digit with even digit is 4*5*5*5 =500 (the first digit could be 2,4,6,8 while the others digit could be 0,2,4,6,8)
to be divisible by eight the last 3 digits must be divisible by 8
so the first digit could be any number hence it is 4
if the second digit is 0 and the third digit is 0 then the last digit could be only 8
if the second digit is 0 and the third digit is 2 then the last digit could be only 4
if the second digit is 0 and the third digit is 4 then the last digit could be only 8
if the second digit is 0 and the third digit is 6 then the last digit could be only 4
if the second digit is 0 and the third digit is 8 then the last digit could be only 8
if the second digit is 2 and the third digit is 0 then the last digit could be only 8
if the second digit is 2 and the third digit is 2 then the last digit could be only 4
4 8
6 4
8 8

Hence it seems that the second digit will have 5 set that can be divided by 8
hence we have0,2,4,6,8 --> 5*5*5

hence prob is 1/4

(A) 1/8
(B) 3/16
(C) 1/4
(D) 8/25
(E) 2/5

a no is divisible by 8 if last 3 digits are divisible by 8
total possibilities =5*5*5(even no)=125
no divided by 8 = 40
probaibility=40/125=8/25

The Answer is E -2/5
To find the probability that the selected number is divisible by 8, we need to count the total number of valid four-digit positive integers with all even digits and then count the number of integers among them that are divisible by 8. For each digit in the four-digit number, we have five choices (0, 2, 4, 6, or 8), except for the first digit, which cannot be 0. Thus, there are
5×5×5×5 valid four-digit positive integers with all even digits.
Next, we need to count how many of these valid integers are divisible by 8. A number is divisible by 8 if the last three digits form a multiple of 8. Since all the digits are even, the last digit must be 0, 2, 4, 6, or 8. This gives us five possible choices for the last digit.
Now, for the remaining three digits (the first three digits), any combination of even digits can be used, including repetition. There are five even digits available (0, 2, 4, 6, and 8). Therefore, the total number of valid four-digit integers that are divisible by 8 is
5×5×5
Now, we can calculate the probability:
So, the correct answer is option (E): 2/5.

We can see that 1 in every 4 positive integers that have all their digits as even numbers will be divisible by 8
-> The probability is 1/4

Answer C

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

Total case = 4*5*5*5 = 500 (First digit can be 2,4,6,8 and 2nd digit onwards can be 0,2,4,6 & 8)

favourable case = 160

probability= 160/500 = 8/25

Option D is correct

Given: A number is selected at random from four-digit positive integers that have all their digits as even numbers.
Asked: What is the probability that the selected number is divisible by 8?

Total numbers with all their digits as even numbers = {2000, 2002, 2004, 2006, 2008, 2020, 2022, 2024, 2026, 2028, 2040, 2042, 2044, 2046, 2048, 2060, 2062, 2064, 2066, 2068, 2080, 2082, 2084, 2086, 2088, ...} = 25*20
Total number with all their digits as even numbers and divisible by 8 = {2000, 2008, 2024,2040, 2048, 2064, 2080, 2088, ...} = 8*20

Total 4-digit numbers with all their digits as even numbers = 4*5*5*5 = 500 = 25*20

The probability that the selected number is divisible by 8 = 8*20/25*20 = 8/25

IMO D

Total number of 4 digit numbers with even digits: 4*5*5*5: 4 accounts for 2,4,6,8 and 5 accounts for 2,4,6,8,0
Total number of 4 digit numbers with even digits such that it is divisible by 8: 4*(8*5): 4 accounts for 2,4,6,8 and 8*5 accounts for different numbers for every even numbers (2,4,6,8,0)
hence probability: 8*5*4/20*25=8/25. D is the answer.

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 8/25
(E) 2/5

Solution:

Denominator:
All are four digit even nos. So the four places can be filled by 0,2,4,6,8.
_ _ _ _ - These 4 places can be filled in 4*5*5*5 ways.

Numerator:
In 1 hundred set, we have 8 multiples of 8 that are purely made of even nos. - for eg: 000, 008, 024, 040, 048, 064, 080, 088.
These 8 will repeat alternately in numbers greater than 2000, 4000, 6000, 8000.
In numbers greater than 2000 we have total 8*5 such nos.
So, total will be 8*5*4 (to include all nos. greater than 2000, 4000, 6000, 8000).

Numerator/Denominator = 8*5*4/4*5*5*5 = 8/25. ANSWER D

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

(A) 1/8
(B) 3/16
(C) 1/4
(D) 8/25
(E) 2/5

For any number to be divisible by 8, last 3 digits should be divisible by 8
total number of four-digit integers that have all their digits as even numbers = 4 *5*5*5=500

in every 2000 to 2888, we will have only 40 digits divisible by 8
2000, 2002, 2004, 2006, 2008, 2020, 2022, 2024, 2026, 2028, 2040, 2042, 2044, 2046, 2048, 2046, 2062, 2064, 2066, 2068, 2080, 2082, 2084, 2086, 2088
it has 8 digits divided by 8
2200, 2202, 2204, 2206, 2208, 2220, 2222, 2224, 2226, 2228, 2240, 2242, 2244, 2246, 2248, 2246, 2262, 2264, 2266, 2268, 2280, 2282, 2284, 2286, 2288
by symmetry it has 8 as well
Similarly between 2400 to 2488
similarly between 2600 to 2688
similarly between 2800 to 2888
and by symmetry we will have 40 for four digit starting with 2 *4 (2,4,6,8 four thousand digits)= 160 digits divisible by 8
so total probability = 160/500 = 8/25

D is the answer

A tough one !

Number of four-digit positive integers that have all their digits as even numbers = 4 * 5 * 5 * 5

The thousands place can be filled in 4 ways (2 / 4/ 6 / 8), and all other places can be filled in 5 ways (0/2/4/6/8)

A number is divisible by 8 when the last three digits are divisible by 8.

A B C D

1000*A + 100 * B + 10 * C + D

B can be (0/2/4/6/8) , hence 100*B is always divisible by 8.

We need to concentrate only on the last two digits

00
02
04
06
08

20
22
24
26
28

40
42
44
46
48

60
62
64
66
68

80
82
84
86
88

8 applicable numbers.

Favorable numbers :

The last two digits can be filled in 8 ways , the thousands place can be filled in 4 ways and the hundred place can be filled in 5 ways

Probability = (4 * 5 * 8) / (4 * 5 * 5 * 5) = 8 / 25

IMO D

Given : A number is selected at random from four-digit positive integers that have all their digits as even numbers.

To find : Probability that the selected number is divisible by 8.

Let abcd be the 4 digit number.

a can be any one of 2,4,6,8
b,c,d can be any of 0,2,4,6,8

Total 4 digit numbers with all even positive integers = 4 * 5 *5 * 5 = 500

lets calculate the number of 4 digit numbers with all even positive integers that are divisible by 8.
For a number to be divisible by 8, the last three digits need to be divisible by 8.
Also since the digits needs to be only even , we can't apply AP logic.
So manually finding out the value that 'bcd' out of 'abcd' can take.
Since b will be even , it can be 0,2,4,6,8, notice that 000, 200, 400,... all will be divisible by 8. So, we can only concentrate on finding combinations for cd first and later include (b*100) in that .

cd can be any one of - 00 ,08 ,24, 40 , 48, 64, 80, 88 (Total 8 combinations)
b can take any of the value out of 0,2,4,6,8 (Total 5 combinations)
a can take any of the value out of 2,4,6,8 (Total 4 combinations)

Total number of 4 digit numbers with all even positive integers that are divisible by 8 = 8 * 5 * 4 = 160

required Probability = 160/500 = 8/25

Answer : (D) 8/25

A number is selected at random from four-digit positive integers that have all their digits as even numbers. What is the probability that the selected number is divisible by 8?

divisibility rule for 8:- if last 3 digit number is divisible by 8, then the number is divisible by 8.

since all the numbers are even, set of even digit = [0,2,4,6,8]

lets abcd be 4 digit number, in which a, b, c and d are even digit
Now since it is a 4 digit number first number cannot be zero.

a cannot be equal to 0, hence total number of ways of selecting 1 digit from 4 digit [2,4,6,8] =4c1 = 4

we need to select bcd in a way that all the digit are even and divisible by 8,

We will check the number divisible from 8 from 0 to 200, (why 200 and not 100 is because we have to understand the pattern, and 0 to 200 will give you the right pattern instead of 0 to 100)
bcd =
[000,008, 024, 040,048,064]
[080, 088]

So total 8 digit from 0 to 200 that are divisible by 8 and has even digits.

From 0-1000, we need to multiple that number 5, so total three digit having even digits that are divisible by 8 = 8*5

so a has 4 options and bcd total options are 8*5

so total number of 4 digit number having even digit that is divisible by 8 = 4*8*5

Total number of even digit numbers having even digit = 4*5*5*5

to probability = 4*8*5/(4*5*5*5) = 8/25

Hence answer is
(D) 8/25