number arrangements help (1 Viewer)

[darkphoenix]

number less than 4000 using 1,3,5,8,9
how many of them are divisible by 3

I did this question in a long way, using that the sum of digit equals is divisible by 3 like 1359, there is 2*3! number which is divisible by 3. can someone show me a short method for this? Thanks

 

[IamBread]

 

[Shadowdude]

number less than 4000 using 1,3,5,8,9
how many of them are divisible by 3

I did this question in a long way, using that the sum of digit equals is divisible by 3 like 1359, there is 2*3! number which is divisible by 3. can someone show me a short method for this? Thanks

That question doesn't make sense to me?

So first we form numbers with 1, 3, 5, 8 and 9 - that are less than 4000 and then we count how many are divisible by three?

 

[IamBread]

That question doesn't make sense to me?

So first we form numbers with 1, 3, 5, 8 and 9 - that are less than 4000 and then we count how many are divisible by three?

I wasn't to sure, so I went with "How many numbers below 4000 are divisible by 3"

 

[darkphoenix]

That question doesn't make sense to me?

So first we form numbers with 1, 3, 5, 8 and 9 - that are less than 4000 and then we count how many are divisible by three?

Yea

 

[Shadowdude]

Repetitions of numbers allowed, or not allowed?

 

[darkphoenix]

Oops forgot to mention , no repetition

 

[D94]

Is the answer 71? If not, what's the answer?

 

[darkphoenix]

The answer is 50.

 

[GingerKid1]

Fail......... newbs

 

[darkphoenix]

for 1 digit numbers theres 2 numbers divisible 3
for 2 digit numbers its 4*2 since last 2 digits has to be either 3 or 9
for 3 digit numbers its 4*3*2 since last 2 digits has to be either 3 or 9 thus 4 digits will be remaining
for 4 digit numbers they have to start with either 1 or 3 and end with 3 or 9 therfore 2*3*2*2 since

so when you add em up its 58 though ans is 50 apparently:/ sum1 tell em where i went wrong

well, last digit with 3 or 9 may not be divisible by 3, for example, 13 is not, nor 19... the only method i can find is what i did, for example the sum of digit 138 is 12 which is divisible by 3, there for, all arrangement of 138 is divisible by 3... like 381,813 etc. But this method is really slow when trying to find the actually sum of digit which is divisible by 3, that's why i am searching for faster method

 

[zeebobDD]

well, last digit with 3 or 9 may not be divisible by 3, for example, 13 is not, nor 19...]

ohhh fk biggest mistake ever:L forgot my divisibility rules hahah fuamaad