Elitmus
Exam
Numerical Ability
Arithmetic
#Elitmus 30th Nov,2014 question.
How many 3 digits numbers are there whose
product of digits is 36? Options :
a)12
b)15
c)18
d)21
Read Solution (Total 6)
-
- x*y*z = 36
where x,y,z are digits of number < 10
(1,4,9),(1,6,6),(2,2,9),(2,3,6),(3,3,4) & its permutation.
6 + 3 + 3 + 6 + 3 = 21 numbers - 10 years agoHelpfull: Yes(47) No(1)
- No. of combinations possible are:
(4,3,3) - 3!/2! ways=3 ways
(2,6,3) - 3! ways=6 ways
(2,2,9) - 3!/2! ways=3 ways
(4,9,1) - 3!ways=6 ways
(1,6,6) - 3!/2! ways=3 ways
total= 3+6+3+6+3=21 ways - 10 years agoHelpfull: Yes(14) No(0)
- option d 21
factor of 36=2*2*3*3
five basic no are 149,166,229,236,334 whoose product of digit is 36
eg 149 rotate in 6 form 149,194,941,914,419,491
236 rotate i also 6 form 236,326,362,263,623,632
now 166 in 3 form 166,616,661
229 in 3 form 229,292,922
334 in 3 form 334,343,433
so total combination is 6+6+3+3+3=21 - 10 years agoHelpfull: Yes(5) No(1)
- answer is 21
- 10 years agoHelpfull: Yes(1) No(0)
- i think 21 times..is it correct
- 10 years agoHelpfull: Yes(1) No(1)
- (2,2,9) -> total 3!/2=3
(1,4,9) -> 3!=6 as no repetition
(3,3,4) -> 3!/2=3
(1,6,6) -> 3!/2=3
(2,3,6) -> 3!=6
so total 3 digit no= 3+6+3+3+6=21 - 10 years agoHelpfull: Yes(0) No(0)
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