Latest Elitmus Aptitude Question SOLUTION: A seven-digit number comprises of only 2's and 3's. How many of these are multiples `of 12? (A)1 (B)11 (C)21 (D)44

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(#M40037561) ELITMUS Numerical Ability question number sys Keep an EYE

A seven-digit number comprises of only 2's and 3's. How many of these are multiples `of 12?
(A)1
(B)11
(C)21
(D)44

Asked In Elitmus Mukesh Nayal (9 years ago)

Unsolved Read Solution (16)

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Read Solution (16) :

RAKESH  (9 years ago)

(B) 11
for a number to be divisible by 12, it should be divisible by 3 and 4 both.
last two digit will be 32 only.
-----32
sum of digits should be multiple of 3, last two digit sum is 5 so first 5 digit sum can be 10,13 only.
so, 1st 5 digit can be 22222 or 22333 .

numbers are 2222232 , 2233332 & its permutation.
total no.s = 1 + 5!/(3!*2!) = 11

Is this solution Helpfull? Yes (34) | No (2)  

Sakshi kochhar  (9 years ago)

for the number to be divisible by 12 it has to be divisible by 4 and 3 both
1. to be divisible by 4 last two digits have to be 32
-----32
2. to be divisible by 3
case 1 : all digits be 2 i.e 2222232
case 2 : three 2's and two 3's be arranged in 5!/(2!*3!) = 10
case 3 : two 3's and three 3's be arranged in 5!/(2!*3!) =10
therefore total number of ways =1+10+10=21

Is this solution Helpfull? Yes (9) | No (16)  

Shradha Agarwal  (9 years ago)

11 solutions are available. 3332232, 3323232, 3233232, 2333232, 3322332,3223332, 2233332, 3232332, 232332, 2332332, 2222232

Is this solution Helpfull? Yes (6) | No   

Mukesh Nayal  (9 years ago)

@ Sonu 3332232 is also possible .There will be many more.

Is this solution Helpfull? Yes (3) | No (1)  

sonu sahoo  (9 years ago)

again calculation mistake its 10+1=11

Is this solution Helpfull? Yes (3) | No   

sunita  (9 years ago)

Answer is 11 only...
for divisibility by 4...last two digit has to be 32...
now seven digits look like -----32
case1: When all first five digits are 2
case 2:when we have two 2s and three 3's coz it makes sum 18 so divisible by 3
this will be 5!/(2!*3!)=10
therfore 10+1=11

Is this solution Helpfull? Yes (3) | No (1)  

sonu sahoo  (9 years ago)

only 1 such number is possible...2222232 because factors of 12=3*4 last two digits must be divisible by 4 so it happens only if last two digits are 32 and for 3 sum of all digit is divisible by 3 so 2spots are already filled rest are 5 ,sum of 3+2=5 i.e sum of first five digit is either 10 or 13 so sum become 15 or 18 ,second case is not possible , first case is possible if first five digits are 2's .....

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Sakshi kochhar  (9 years ago)

sorry case 3 will not be there
ans is 1+10=11 only

Is this solution Helpfull? Yes (2) | No   

ruby gupta  (9 years ago)

11 is ryte ans
coz only 2 cases r possile
1-2222232
2-2233332
n in case 2 we arrange the no by its permutation
5!/3!*2!=10
then total caes =10+1=11

Is this solution Helpfull? Yes (1) | No   

sonu sahoo  (9 years ago)

21 is the ans , first num is 2222232, 3332232 is also possible as last two digits are fixed so first five digits can arrange them in 5!/3!*2!=20 , 20+1=21 ,

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pankaj  (9 years ago)

22222 = 1
two(2)and three(3)=10
three(2) and three 3=10

total 21 answer

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Prakash Singh  (9 years ago)

I think 21 is correct answer...

Is this solution Helpfull? Yes | No (1)  

lalita kesarwani  (9 years ago)

ans. is 21 as there are 2 cases case 1. when no. ends with 21 nd case 2. when no. ends with 22 now for its divisibility by 3 there are another 2 cases when sum is either 15 or 18

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Navpreet Singh  (9 years ago)

solution are the form of
-----22
then 1st five position filled by
sum of first five =13
then the only combination =2 3's and 3 2's
the permuatution are
5!/3!2!=10
and one combination is 2222232
so total=1+10=11 ans

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Navpreet Singh  (9 years ago)

solution are the form of
-----22
then 1st five position filled by
sum of first five =14
then the only combination =2 3's and 3 2's
the permuatution are
5!/3!2!=10
and one combination is 2222232
so total=1+10=11 ans

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neelaksh  (7 years ago)

Answer is 11
Number should be a multiple of 3 and 4.
So, the sum of the digits should be a multiple of 3.

We can either have all seven digits as 3, or have three 2's and four 3's, or six 2's and a 3 (The number of 2's should be a multiple of 3).

For the number to be a multiple of 4, the last 2 digits should be 32. Now, let us combine these two.

(i)- All seven 3's - No possibility

(ii)- Three 2's and four 3's - The first 5 digits should have two 2's and three 3's in some order. No of possibilities =
5!
3!×2!
=10

(iii)- Six 2's and one 3 - The first 5 digits should all be 2's.
So, there is - only one number - 2222232

So we get the total of seven-digit number comprises of only 2's and 3's which are multiples of 12 are:
= (i) + (ii) + (iii)
= 0+10+1=11

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