1057the first question...
if its only 5 digit numbers then....
notice that 4x4! numbers can be formed using 0,2,4,6,8....
_ _ _ _ _
note in the ten thousandth place each of 2,4,6,8 occur 4! times and in the remaining places each occur 3x3! times...
so sum is (2+4+6+8)(4x4!x104+3x3!(103+102+10+1))
for the second question there are 5!x3 even numbers possible....out of this u have to find which are divisible by 3.....since in the set there are equal number of numbers of the from 3n,3n+1,3n+2....in all also there should be equal number of numbers....
so total numbers divisible by 6 is 5!x33 = 5!=120
1ketan , answer for 1st one isnt matching . i think there mistake in last step . well thanks for approach .
also plz check ,
i did the 2nd one as finding the number of multiples of 6 in the smallest and largest number formed by given numbers and then subtracting them to get ans . but it doesnt comes 120 ..... whats wrong here??
1@Ketan, shouldn't the answer be 360 only? There are 3*5! even numbers and the sum of all the digits is divisible by three. so any even number should be divisible by 6
Am I correct? Can you spot my mistake?
1057
Ketan Chandak
·2012-03-30 08:53:43
u are right.....dun knw wat i was thinking.... :/
1
mast
·2012-03-30 09:20:17
thanks blueivy , didnt thought that .
