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How many five-digit positive integers can be formed using each of the 08 May 2020, 09:30
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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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E
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How many five-digit positive integers can be formed using each of the Updated on: 08 May 2020, 17:46
Originally posted by sujoykrdatta on 08 May 2020, 16:58.
Last edited by sujoykrdatta on 08 May 2020, 17:46, edited 1 time in total.
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Bunuel
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
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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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How many five-digit positive integers can be formed using each of the 08 May 2020, 17:34
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sujoykrdatta
Bunuel
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
Answer C
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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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How many five-digit positive integers can be formed using each of the 08 May 2020, 17:40
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sujoykrdatta
Bunuel
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
Answer C

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
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Oops - Yes :|

I guess I am making school-level errors now :|

Corrected - thank you
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How many five-digit positive integers can be formed using each of the 31 May 2020, 02:35
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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.
Answer (E).
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How many five-digit positive integers can be formed using each of the 10 Jul 2021, 21:40
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Asked: 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?

15 = 3*5
The 5-digit number formed using each of the digits from {1, 2, 3, 5} at least once such that they are a multiple of 15 should be multiple of 3 & 5 both.

For the number to be multiple of 5, last digit should be 5 only since 0 is not available.
For the number to be multiple of 3, sum of digits should be multiple of 3.

The 5-digit number be abcd5.
a + b + c + d + 5 = 3k
a + b + c + d = 3k - 5 = 3(k-2) + 1 = 3m + 1

Since each of the digits from {1, 2, 3, 5} is used at least once.
1 + 3 + 2 + 5 = 11
11 + 1 = 12

Therefore, 1 will be used twice while other digits from {2,3} is will be used once and 5 will be the last digit.

Number of such cases = 4!/2! = 12

IMO E
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How many five-digit positive integers can be formed using each of the 10 Jul 2021, 21:44
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Hi sujoykrdatta

You can not use numbers 4 & 7 since the five-digit positive integers can be formed using each of the digits from {1, 2, 3, 5} at least once.

sujoykrdatta
Bunuel
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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How many five-digit positive integers can be formed using each of the 16 Jan 2024, 08:52
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Bunuel
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
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Multiple of 15 means the number is multiple of 3 and 5 too. If the number is multiple of 3, then the sum of digits is a multiple of 3. If the number is multiple of 5, then the number is divisible by 5 which means last digit shall always be 5. Now, 1+2+3+5=11. If we add 1 to it, then the sum becomes 12 and it becomes a 5-digit number too. Arrange the 5 digits keeping 5 at the end - 4!/2!=12(E).
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How many five-digit positive integers can be formed using each of the 16 Apr 2025, 08:19
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