12 Days of Christmas GMAT Competition - Day 8: How many 5-digit

Question

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How many 5-digit positive integers, where each digit is a prime number, are divisible by 25?

A. 64
B. 96
C. 128
D. 192
E. 256

Bunuel · Accepted Answer

GMAT Club's Official Explanation:

For a number to be divisible by 25, it must end in 00, 25, 50, or 75. Since we are looking for numbers composed entirely of prime digits, only 25 and 75 are valid options for the last two digits. Thus, there are two possibilities for the last two digits.

The prime digits are 2, 3, 5, and 7, meaning each of the first three digits can independently take any of these four values. This gives a total of 4 * 4 * 4 = 64 combinations for the first three digits.

Therefore, the total number of five-digit positive integers, where each digit is a prime number and the number is divisible by 25, is 2 * 64 = 128.

Answer: C.

GraCoder · Answer

Notice that 100x + y(y < 100) would leave a remainder of y, hence y can only be 25, 75, 50, 00 to be divisible by 25.

For this ques the only possibilities for last two digits are 25 and 75

for the remaining 3 digits we have 4*4*4 choices => 64 for each 25 and 75 => 128 total ways.

Krunaal · Answer

Each of the 5 digits is a prime number, therefore each of the 5 digits can be from {2, 3, 5, 7}

We have to find how many of them are divisible by 25, there are two cases where last two digits can be either 25 or 75.

_ _ _ 2 5 or _ _ _ 7 5

The three _ _ _ digits can be from the 4 prime numbers {2, 3, 5, 7} = 4*4*4 = 64 possibilities for each case.

For both cases = 64 + 64 =  128

Answer C.

hr1212 · Answer

Single digit prime numbers - 2,3,5,7

Last 2 digits which are divisible by 25 => 25 and 75

Possible combinations:

First digit 
 Second digit 
 Third digit 
 Fourth digit 
 Fifth digit 
 
  4 (All 4 digits are feasible) 
 4 (All 4 digits are feasible) 
 4 (All 4 digits are feasible) 
 2 (Either 2 or 7 is feasible) 
 1 (Only 5 is possible)

Multiplying all combinations = 4*4*4*2 = 128
Answer: C

ManifestDreamMBA · Answer

Since, each digit is prime, we can only use 2,3,5,7. For a number to be divisible by 25, the last 2 digits need to be 25 or 75.

By slot method for 5 digits, left to right
Digit 1 = 4 options/no restriction
Digit 2 = 4 options/no restriction
Digit 3 = 4 options/no restriction
Digit 4 = 2 options - 2 or 7
Digit 5 = 1 option, only 5

we have 4*4*4*2*1 = 128 values

AbhiS101 · Answer

Prime digits:  The prime digits are 2, 3, 5, and 7.
 Divisibility by 25:  A number is divisible by 25 if its last two digits are 00, 25, 50, or 75. Since we can only use prime digits, the last two digits must be 25, 50, or 75. However, since 0 is not a prime number, only 25 and 75 are possible.
 First three digits:  The first three digits can be any of the four prime digits (2, 3, 5, or 7). Therefore, there are 4 choices for each of the first three digits.
 Counting the numbers: 
If the last two digits are 25, there are 4 * 4 * 4 = 64 possible numbers.
If the last two digits are 75, there are 4 * 4 * 4 = 64 possible numbers.
 Total:  The total number of 5-digit integers is 64 + 64 = 128

IMO C

A_Nishith · Answer

Step 1:  Understand the conditions
Prime digits: The digits must be selected from the set of prime numbers:{2,3,5,7}

Divisibility by 25: A number is divisible by 25 if its last two digits form a number that is divisible by 25. 
Thus, the possible combinations for the last two digits are: 25,75
(Since 2, 3, 5, and 7 are the only valid digits, only these two combinations work.)

Step 2:  Form a 5-digit number
A 5-digit number has the following structure: d1 d2 d3 d4 d5

Where:
d1 , d2 , and d3 can be any prime digit from {2,3,5,7}.
d4 d5 must be either 25 or 75 for divisibility by 25.

Step 3 : Calculate the total number of such numbers
Choices for d1 , d2 , and d3 : Each of these three digits can independently take on any of the 4 values {2,3,5,7}.

Thus, there are: 4×4×4=64 possible combinations for d1 d2 d3

Choices for d4 d5 : The last two digits can be either 25 or 75. 
Thus, there are 2 choices.

Total combinations: Multiply the choices for d1 d2 d3 by the choices for d4 d5 : 64×2=128

Final Answer: 
The total number of 5-digit integers meeting the criteria is:  128

nishantswaft · Answer

The stem tells us that;

5-digit positive integers   
  each digit is a prime number

And we have to select;  
 
  How many are divisible by 25?

A number is divisible by 25 only when its last two digits are divisible by 25. 
 Since there are only prime-numbered digits; 
 Last two digits can only be 25 or 75.

_ _ _ _ _

Let's start from the place value with most constraints.

The last digit can only be 5 so possible values =1
The second last digit can only be 2,7 so possible values =2 
The middle digit can only be 2,3,5,7 so possible values =4 
The second digit can only be 2,3,5,7 so possible values =4 
The first digit can only be 2,3,5,7 so possible values =4

Total values= 4*4*4*2*1= 128

twinkle2311 · Answer

Prime digits are 2, 3, 5, and 7.

For divisibility by 25, the last two digits must be 25, 50, 75, or 00.

Since only 2, 3, 5, and 7 are allowed, valid pairs are 25 and 75. Fix the last two digits as 25 or 75.

The first three digits can each be 2, 3, 5, or 7.

For each case:
 
 4 choices for each of the first three digits → 4 × 4 × 4 = 64. 
 Two valid endings (25 or 75) → 64 × 2 = 128.

Answer: C

crimson_king · Answer

For a 5 digit number we need to ensure that the last two digits are either 25,50,75 & 00. We can eliminate 50 & 00 as all the digits are required to be prime numbers.

Hence we are left with only those options that have last two digits 25 & 75. The first three digits are also prime numbers & repetition is allowed, hence the total number of possibilities are 4*4*4=64 numbers for one possible last two digit option of 25.

For 75 as well the total number of possibilities are again 4*4*4=64 numbers.

So the total number of possibilities are - 64*2=128 possibilities.

Hence the answer is option (C) 128

Karanjotsingh · Answer

Solve this step by step: 
 
 First, let's identify what we need:
 
 5-digit numbers (10000 to 99999) 
 Each digit can only be prime (2, 3, 5, 7) 
 Must be divisible by 25  
 
 Key insight:
 
 For a number to be divisible by 25, its last two digits must form a number divisible by 25 
 Looking at the last two digits, using only 2,3,5,7:
 
 Only 25 and 75 are possible as the last two digits

Now for our count:
 
 First 3 digits: can be any prime (2,3,5,7) 
 The last 2 digits: must be either 25 or 75 
 So we have: 4 × 4 × 4 × 2 possibilities 
 = 64 × 2 
 = 128

Answer: C. 128

pintukr · Answer

How many 5-digit positive integers, where each digit is a prime number, are divisible by 25?

A. 64
B. 96
C. 128
D. 192
E. 256

For divisibility by 25, last two digits should be 00 / 25 / 50 / 75

Since, each digit is a prime number , only 25 / 75 are the possibilities for ten's digit and unit's digits, respectively.

Case 1 : 25 for ten's digit and unit's digits

For ten-thousandth, thousandth, and hundredth digits ---> possibilities of 'prime' digits are 2/3/5/7 i.e, 4 possibilities

So, Total possibilities = 4 * 4 * 4 = 64.............(1)

Case 2 : 75 for ten's digit and unit's digits

For ten-thousandth, thousandth, and hundredth digits ---> possibilities of 'prime' digits are 2/3/5/7 i.e, 4 possibilities

So, Total possibilities = 4 * 4 * 4 = 64.............(2)

Using (1) & (2),

Total such 5-digit positive integers = 64 + 64 = 128

(C) is the CORRECT answer

Navya442001 · Answer

The prime digits are 2,3,5,7, so each digit of the number must be one of these values.

A number is divisible by 25 if its last two digits are 00,25,50,75
Only prime digits allowed so, 25 and 75
First 3 digits should be from prime numbers, so, 2,3,5,7
Total number of combinations for first 3 digit = 4x4x4 = 64
Total number of combinations for first 5 digits = 64x2 = C. 128

For numbers ending in 25, the first three digits can independently be any of the prime digits 2,3,5,7
 This gives 4×4×4=64

Similarly, for numbers ending in 75, the first three digits can independently be any of the prime digits 2,3,5,7. 
This also gives 4×4×4=64

Adding the two cases together, the total is 64+64=128

The total number of 5-digit positive integers divisible by 25, where each digit is a prime number, is 128
 Correct answer: C.

Prakhar9802 · Answer

For a 5 digit positive integer, with each digit a prime number, let's make 5 blank spaces -

_ _ _ _ _

Prime numbers with single digits = 2,3,5,7

For any number to be divisible by 25, it's last two digits must be divisible by 25.
So, the last digit must be 5, as 0 is not a prime number.

_ _ _ _ 5 = 1 way

The second last digit can either be 2 or 7

_ _ _ 2,7 5 = 2*1 way

for the remaining three spaces, any of the 4 prime numbers can occupy those spaces = 4*4*4*2*1 ways

therefore, 128 ways.

FINAL ANSWER - Option C

Devaditya · Answer

Prime Digits: The prime numbers less than 10 are 2, 3, 5, and 7. So, each digit of the 5-digit number must be one of these four prime numbers.

Divisibility by 25: A number is divisible by 25 if its last two digits form a number that is divisible by 25. The possible pairs of digits that are prime and form a number divisible by 25 are 25 and 75.

Now, let's count the number of 5-digit integers where each digit is a prime number and the number is divisible by 25:

The last two digits can be either 25 or 75.

The first three digits can each be any of the four prime numbers (2, 3, 5, 7).

So, we have: 4×4×4×2=128
Therefore, the number of such integers is 128

Tishaagarwal13 · Answer

The  correct answer is C

A number will be divisible by 25 if the last two digits are divisible by 25

Since each digit is a prime number, possible numbers can be:
_ _ _ 25 and _ _ _ 75

The first three digits can be any of the prime digits between 2,3,5,7.

And, we are not given that the digits are distinct.
So, the first digit can have 4 options, and the second and third digits can also have 4 options each

_ _ _ 25 - 4*4*4 = 64 
_ _ _ 75 - 4*4*4 = 64

Total = 128

IssacChan · Answer

A single prime number digit include 2,3,5,7 only

To be divisible by 25, the last 2 digit should be either 25 or 75.
So 5th digit can only be 5 - 1 choice only
4th digit can either be 2 or 7 - 2 choices
1st to 3rd digit can be either 2,3,5,7 - 4 choices for each digit

So the number of 5 digit positive integers can be 4(1st digit)*4(2nd digit)*4(3rd digit)*2(4th digit)*1*(5th digit)
= 128
C is the answer

Mardee · Answer

A number is divisible by 25 if its last two digits are 00,25,50,75
Only prime digits allowed so, 25 and 75
First 3 digits should be from prime numbers, so, 2,3,5,7
Total number of combinations for first 3 digit = 4x4x4 = 64
Total number of combinations for first 5 digits = 64x2 =  C. 128

prashantthawrani · Answer

How ma ny 5-digit positive integers, where each digit is a prime number, are divisible by 25?

Since 5 digit integer is divisible by 25, last two digits must be 25 or 75.

__ __ __  2 or 7   5

First 3 positions have 4 possible values=2,3,5,7
4th position has 2 possible values=2,7
5th position has 1 possible values=5

No. of possible integers= 4*4*4*2*1=128

C. 128

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First digit	Second digit	Third digit	Fourth digit	Fifth digit
4 (All 4 digits are feasible)	4 (All 4 digits are feasible)	4 (All 4 digits are feasible)	2 (Either 2 or 7 is feasible)	1 (Only 5 is possible)

GMAT Problem Solving (PS) Questions

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Prep Toolkit

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