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Math Expert V
Joined: 02 Sep 2009
Posts: 64997
How many five-digit positive integers can be formed using each of the  [#permalink]

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Difficulty:   55% (hard)

Question Stats: 55% (02:03) correct 45% (02:49) wrong based on 42 sessions

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How many five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15?

A. 36
B. 24
C. 18
D. 15
E. 12

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How many five-digit positive integers can be formed using each of the  [#permalink]

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Bunuel wrote:
How many five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15?

A. 36
B. 24
C. 18
D. 15
E. 12
Are You Up For the Challenge: 700 Level Questions

Note: The question says that you should use 1, 2, 3, 5 at least once; it does not say that you should use only those digits

To make a number a multiple of 15, the number must be a multiple of 5, which, in this case, only possible if the units digit is 5 or 0
Also, the number must be a multiple of 3 => Sum of the digits is a multiple of 3

Since we need to use all the digits at least once, we already have a 4-digit number with sum of digits 1+2+3+5 = 11

If we wish to use the digit 0 at the end (units digit), the digits would be 1, 2, 3, 5, 0 => sum of the digits is 11, not divisible by 3. Hence, 0 cannot be used.

Thus, the possibilities are (keeping 5 fixed at the end):

# 1, 2, 3, 1, 5 => sum = 12 => Number of ways = 4!/2! = 12
# 1, 2, 3, 4, 5 => sum = 15 => Number of ways = 4! = 24
# 1, 2, 3, 7, 5 => sum = 18 => Number of ways = 4! = 24

Thus, we can form 12 + 24 + 24 = 60 such numbers

Now: This does not match any of the answer options.
So I am thinking that the question actually wanted to say that you can use ONLY the digits 1,2,3,5 at least once.
In that case, the answer is 12 (option E)
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Sujoy Kumar Datta
Director - CUBIX Educational Institute Pvt. Ltd. (https://www.cubixprep.com)
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Originally posted by sujoykrdatta on 08 May 2020, 15:58.
Last edited by sujoykrdatta on 08 May 2020, 16:46, edited 1 time in total.
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Re: How many five-digit positive integers can be formed using each of the  [#permalink]

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sujoykrdatta wrote:
Bunuel wrote:
How many five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15?

A. 36
B. 24
C. 18
D. 15
E. 12
Are You Up For the Challenge: 700 Level Questions

Note: The question says that you should use 1, 2, 3, 5 at least once; it does not say that you should use only those digits

To make a number a multiple of 15, the number must be a multiple of 5, which, in this case, only possible if the units digit is 5 or 0
Also, the number must be a multiple of 3 => Sum of the digits is a multiple of 3

Since we need to use all the digits at least once, we already have a 4-digit number with sum of digits 1+2+3+5 = 11

If we wish to use the digit 0 at the end (units digit), the digits would be 1, 2, 3, 5, 0 => sum of the digits is 11, not divisible by 3. Hence, 0 cannot be used.

Thus, the possibilities are:

# 1, 2, 3, 5, 1 => sum = 12
# 1, 2, 3, 5, 4 => sum = 15
# 1, 2, 3, 5, 7 => sum = 18

In each case, since the units digit is 5, we can arrange the other 4 digits (with one digit repeating) in 4!/2! = 6 ways
Thus, total cases = 6 x 3 = 18

Thus, we can form 18 such numbers

Hi sujoykrdatta

In the highlighted cases, there is no repetition of digits so the four digits excluding 5 can be arranged in 4! ways right?

Btw, 4!/2!=12 and not 6

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GMAT Tutor B
Status: Entrepreneur | GMAT, GRE, CAT, SAT, ACT coach & mentor | Founder @CUBIX | Edu-consulting | Content creator
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Re: How many five-digit positive integers can be formed using each of the  [#permalink]

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firas92 wrote:
sujoykrdatta wrote:
Bunuel wrote:
How many five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15?

A. 36
B. 24
C. 18
D. 15
E. 12
Are You Up For the Challenge: 700 Level Questions

Note: The question says that you should use 1, 2, 3, 5 at least once; it does not say that you should use only those digits

To make a number a multiple of 15, the number must be a multiple of 5, which, in this case, only possible if the units digit is 5 or 0
Also, the number must be a multiple of 3 => Sum of the digits is a multiple of 3

Since we need to use all the digits at least once, we already have a 4-digit number with sum of digits 1+2+3+5 = 11

If we wish to use the digit 0 at the end (units digit), the digits would be 1, 2, 3, 5, 0 => sum of the digits is 11, not divisible by 3. Hence, 0 cannot be used.

Thus, the possibilities are:

# 1, 2, 3, 5, 1 => sum = 12
# 1, 2, 3, 5, 4 => sum = 15
# 1, 2, 3, 5, 7 => sum = 18

In each case, since the units digit is 5, we can arrange the other 4 digits (with one digit repeating) in 4!/2! = 6 ways
Thus, total cases = 6 x 3 = 18

Thus, we can form 18 such numbers

Hi sujoykrdatta

In the highlighted cases, there is no repetition of digits so the four digits excluding 5 can be arranged in 4! ways right?

Btw, 4!/2!=12 and not 6

Posted from my mobile device

Oops - Yes I guess I am making school-level errors now Corrected - thank you
_________________
Sujoy Kumar Datta
Director - CUBIX Educational Institute Pvt. Ltd. (https://www.cubixprep.com)
IIT Kharagpur, TU Dresden Germany
GMAT - Q51 & CAT (MBA @ IIM) 99.98 Overall with 99.99 QA
_________
Feel free to talk to me about GMAT & GRE | Ask me any question on QA (PS / DS) | Let's converse!
Skype: sk_datta, Alt. Email: sujoy.datta@gmail.com
Director  D
Joined: 25 Jul 2018
Posts: 731
Re: How many five-digit positive integers can be formed using each of the  [#permalink]

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How many five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15?

Any five-digit positive integer, which is a multiple of 15 and formed using the digits from {1, 2, 3, 5} should end ....5.
Well, we could form 5-digit numbers:
---> 1123|5: ---> $$\frac{4!}{2!}= 12$$
---> 1223|5: ---> $$\frac{4!}{2!}= 12$$
---> 1323|5: ---> $$\frac{4!}{2!}= 12$$
---> 1523|5: ---> $$\frac{4!}{2!}= 12$$

So, we need to check those numbers whether they are divisible by 3 or not:
---> 1+1+2+3+5= 12 (divisible by 3)
---> 1+2+2+3+5=13 (not divisible by 3)
---> 1+3+2+3+5= 14 (not divisible by 3)
---> 1+5+2+3+5= 16 (not divisible by 3)

In total, there are 12 numbers, which are multiple of 15. Re: How many five-digit positive integers can be formed using each of the   [#permalink] 31 May 2020, 01:35

# How many five-digit positive integers can be formed using each of the  