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A 6-digit number comprises of only 2’s and 3’s. How many of

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A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   23 Oct 2019, 03:22
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A 6-digit number comprises of only 2’s and 3’s. How many of these are multiples of 12?

A. 0
B. 1
C. 6
D. 20
E. 21

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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   23 Oct 2019, 03:39
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Dillesh4096 wrote:
A 6-digit number comprises of only 2’s and 3’s. How many of these are multiples of 12?

A. 0
B. 1
C. 6
D. 20
E. 21

Posted from my mobile device


For it to be a multiple of 12, the number has to be multiple of 3 and 4..
1) Multiple of 4...
Last two digits have to be multiple of 4....Only combination is 32....ABCD32 .. 1 way
2) Multiple of 3..
Since we have used one 2 in 32, we will require to use 2 more 2s in first four digits..
So, ABCD can take two 2s and two 3s in 4!/2!2!=6way

Total 1*6=6

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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   23 Oct 2019, 08:43
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Last two digits should be 32. Rest 4 digits should be 2233 form ----> 4!/(2!*2!) = 6 = answer
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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   23 Oct 2019, 03:40
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Number must be a multiple of 4; hence, last 2 digits must me 32

Number must be a multiple of 3; hence there can be zero, three or six 2's in the number.

Case 1- numbers comprise of zero 2's and six 3's
Such number can't be a multiple of 4

Case 2- numbers comprise of three 2's and three 3's.
Last 2 digits are 32. We can arrange remaining two 2's and two 3's.

Total possible cases= \(\frac{4!}{2!2!}\)=6

Case 3- numbers comprise of six 2's.
such number can't be a multiple of 4

Total possible cases= 0+6+0=6





Dillesh4096 wrote:
A 6-digit number comprises of only 2’s and 3’s. How many of these are multiples of 12?

A. 0
B. 1
C. 6
D. 20
E. 21

Posted from my mobile device
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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   25 Oct 2019, 19:17
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azl wrote:
Hello,

I have a silly question, how do you guys know that the 4 first number must include two 2's and two 3's?

Thank you in advance



Hi,

We have 32 A’s last two digits.
The first 4 can be any of 3 and 2, but all 6 when added should give you a multiple of 3, a property for multiple of 3.
Now you can have as many 3s as they are multiple of 3, but when you have a 2, either you have a 1 with it as 2+1=3 but that is not possible. So you have to have three or six etc of 2. Here you can have three 2s and one 2 is already used in units digit abcd32. So ABCD can have two 2s and the remaining will be 3s

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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink] New post   02 Feb 2022, 11:43
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Re: A 6-digit number comprises of only 2’s and 3’s. How many of [#permalink]
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