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How many positive integers less than 10,000 are such that the product

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How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 02:51
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A
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D
E

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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post Updated on: 20 May 2015, 08:14
1
2
30 = 1 * 30, 2 * 15, 3* 10, 5 *6

We have the option of using single digits 1,2,3,5 and 6
No : of four digit numbers such that the product is 30
The possible combinations are using digits (1,2,3,5) & (1,1, 5, 6) such that the product 30.
No: of different combinations possible using (1,2,3,5) = 4! = 24
No: of different combinations possible using (5,6,1,1) = 4!/2! = 12 ('1' is repeated twice)
Total 4 digit numbers so that the product is 30 is 24+12 = 36

No: of three digit numbers such that the product is 30
The possible combinations are using digits (2,3, 5) & (1,5,6)
No: of different combinations possible using (2,3,5) = 3! = 6
No: of different combinations possible using (1,5,6) = 3! = 6
Total 3 digit numbers so that the product is 30 is 6+6 = 12

No: of 2 digit numbers such that the product is 30 is 2 (56 and 65)

Total numbers under 10000 such that the product is 30 is 36+12+2 = 50


Answer : E

Ambarish
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Originally posted by askhere on 20 May 2015, 03:45.
Last edited by askhere on 20 May 2015, 08:14, edited 1 time in total.
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 07:00
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My attempt:

We need a number from one digit to four digits whose product is 30. None of the digits can be zero then.
The single digit factors of 30 are 1,2,3,5,6 (rest are more than 1 digit)

So we form the combinations as -

Starting with 4 digit numbers
1 _ _ _ 1st combination of 2,3,5 has 3! and 2nd combination of 5,6,1 has 3!
2 _ _ _ combination of 1,3,5 has 3!
3 _ _ _ combination of 1,2,5 has 3!
5 _ _ _ 1st combination of 1,2,3 has 3! and 2nd combination of 6,1,1 has 3!/2!
6 _ _ _ combination of 5,1,1 has 3!/2!

So total 36 ways

Now 3 digit numbers
156 combinations gives 3! ways
235 combination gives 3! ways

So total 12 ways

Now 2 digit numbers
56 combination gives us 2! ways

So total 2 ways

Grand total of ways = 36+12+2=50

I am kind of confused now as none of the options is 50 :)
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 07:39
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askhere wrote:
30 = 1 * 30, 2 * 15, 3* 10, 5 *6

We have the option of using single digits 1,2,3,5 and 6
The possible combinations are using digits (1,2,3,5) & (1,1, 5, 6) to make their product 30.
No: of different combinations possible using (1,2,3,5) = 4! = 24
No: of different combinations possible using (5,6,1,1) = 4!/2! = 12 ('1' is repeated twice)
Total numbers under 10000 so that the product is 30 = 24+12 = 36

Answer : C

Ambarish


hi,
you are forgetting 3 digits and 2 digits number....
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 07:44
1
Bunuel wrote:
How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 40


hello,
we can have 2,3 or 4 digits number satisfying the condition..

4 digits number: it can consist of 1,2,3,5 or 1,1,6,5..
1,2,3,5- 4! ways
1,1,5,6- 4!/2 ways
total-36 ways

3 digits number: it can have 2,3,5 or 1,5,6..
2,3,5 - 3! ways
1,5,6 - 3! ways
total - 12 ways

2 digits- 5,6
total - 2 ways

overall- 50 ways...
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 07:48
Jackal wrote:
My attempt:

We need a number from one digit to four digits whose product is 30. None of the digits can be zero then.
The single digit factors of 30 are 1,2,3,5,6 (rest are more than 1 digit)

So we form the combinations as -

Starting with 4 digit numbers
1 _ _ _ 1st combination of 2,3,5 has 3! and 2nd combination of 5,6,1 has 3!
2 _ _ _ combination of 1,3,5 has 3!
3 _ _ _ combination of 1,2,5 has 3!
5 _ _ _ 1st combination of 1,2,3 has 3! and 2nd combination of 6,1,1 has 3!/2!
6 _ _ _ combination of 5,1,1 has 3!/2!

So total 36 ways

Now 3 digit numbers
156 combinations gives 3! ways
235 combination gives 3! ways

So total 12 ways

Now 2 digit numbers
56 combination gives us 2! ways

So total 2 ways

Grand total of ways = 36+12+2=50

I am kind of confused now as none of the options is 50 :)


Option E should have been 50. Edited.
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 08:16
chetan2u wrote:
askhere wrote:
30 = 1 * 30, 2 * 15, 3* 10, 5 *6

We have the option of using single digits 1,2,3,5 and 6
The possible combinations are using digits (1,2,3,5) & (1,1, 5, 6) to make their product 30.
No: of different combinations possible using (1,2,3,5) = 4! = 24
No: of different combinations possible using (5,6,1,1) = 4!/2! = 12 ('1' is repeated twice)
Total numbers under 10000 so that the product is 30 = 24+12 = 36

Answer : C

Ambarish


hi,
you are forgetting 3 digits and 2 digits number....


Thanks for pointing out. I edited my answer.
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 20 May 2015, 13:53
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Q integers >10,000 are such that the product of their digits is 30

possible factors of 30 that can be digits (5,6,1,1) and (2,3,5,1), so two sets to choose from
Set 1 : (2,3,5,1)
Set 2 : (5,6,1,1)

4 Digit numbers _ _ _ _ (4 spots to be filled by 4 digits )

    Set 1 : (2,3,5,1) _ _ _ _ 4.3.2.1 = 24
    Set 2 : (5,6,1,1) _ _ _ _ (4.3.2.1)/2 = 12

3 Digit numbers _ _ _ (3 spots to be filled by 3 digits )

    Set 1 : (2,3,5) _ _ _ 3.2.1 = 6
    Set 2 : (5,6,1) _ _ _ 3.2.1 = 6

2 Digit numbers _ _ (2 spots to be filled by 2 digits )

    Set 1 : not possible
    Set 2 : (5,6) _ _ _ _ 2.1 = 2

total = 36 + 12 + 2 = 50 Ans : E
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 21 May 2015, 08:00
chetan2u wrote:
Bunuel wrote:
How many positive integers less than 10,000 are such that the product of their digits is 30?


1,1,5,6- 4!/2 ways
...


Correct me if I'm wrong but you have chosen 1,1,5,6 in 4!/2, is it because integer 1 is being repeated twice in set (1,1,5,6)?
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Re: How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 21 May 2015, 18:28
1
RudeyboyZ wrote:
chetan2u wrote:
Bunuel wrote:
How many positive integers less than 10,000 are such that the product of their digits is 30?


1,1,5,6- 4!/2 ways
...


Correct me if I'm wrong but you have chosen 1,1,5,6 in 4!/2, is it because integer 1 is being repeated twice in set (1,1,5,6)?


Hi RudeyboyZ,
you are correct ..
say the set was 1,2,1,1, the ways would have been 4!/3!...here out of 4 positions , there will be 3! will be common .. basically the ways 1 can be placed within themselves...
if the set was 1,1,2,2, the ways would have been 4!/2!2!..
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3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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How many positive integers less than 10,000 are such that the product  [#permalink]

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New post 25 May 2015, 07:21
Bunuel wrote:
How many positive integers less than 10,000 are such that the product of their digits is 30?

A. 12
B. 24
C. 36
D. 38
E. 50


OFFICIAL SOLUTION:

30 = 2*3*5 = 6*5 (only 2*3 gives single digit number 6).

So, we should count the number of positive integers less than 10,000 with the digits {2, 3, 5} and {5, 6} and any number of 1's with each set.

2-digit numbers:
{5, 6} - the number of combinations = 2: 56 or 65.

3-digit numbers:
{1, 5, 6} - the number of combinations = 3! = 6: 156, 165, 516, 561, 615, or 651.
{2, 3, 5} - the number of combinations = 3! = 6.

4-digit numbers:
{1, 1, 5, 6} - the number of combinations = 4!/2! = 12.
{1, 2, 3, 5} - the number of combinations = 4! = 24.

Total = 2 + 6 + 6 + 12 + 24 = 50.

Answer: E.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

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