### Brain Teasers

# Number Trouble

Fun: (2.28)
Difficulty: (2.02)
Puzzle ID: #8587

Submitted By: lesternoronha1 Corrected By: shenqiang

Submitted By: lesternoronha1 Corrected By: shenqiang

Find the lowest positive number divisible by all the single digit numbers 1 to 9.

### Answer

The number is 2520.2*3*2*5*7*2*3=2520

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## Comments

I don't understand how you got the answer.

Piffle, it's very easy.

A number to be divisible by 1, 2, 3,... 9 has to be divisible individually by 2, 3, 4 = 2*2,5,6=2*3,7,8=2*2*2 and 9=3*3.

My answer has 3 2's which means it is divisible by 2,4 and 8; 2 3's which means it is divisible by 3 and 9; and also has a 5 and a 7 which are prime numbers and hence have to be included.

A number to be divisible by 1, 2, 3,... 9 has to be divisible individually by 2, 3, 4 = 2*2,5,6=2*3,7,8=2*2*2 and 9=3*3.

My answer has 3 2's which means it is divisible by 2,4 and 8; 2 3's which means it is divisible by 3 and 9; and also has a 5 and a 7 which are prime numbers and hence have to be included.

I have to agree with Piffle,

This is very confusing but I don't really understand anything as complicated as this. Excellent teaser though.

This is very confusing but I don't really understand anything as complicated as this. Excellent teaser though.

Start with 9 and work backwards. If it is divisible by every number then it must have 9 as a factor. Like-wise 8 since 9 and 8 have no common factor. 9x8=72. Continue with 7 which is prime so 7 must be a factor. 7x72=504. Going backwards to 6 we find that since we have given the number a factor of 9 and 8 then it must be divisible by 6 already. 5 is prime with no common factors so 5x504=2520. Working backwards we have already accounted for 4 with the 8 and 3 with the 9 and 2 with the 8 so it remains 2520.

I did it the same way as Jimbo, working backward from nine. Very easy one, but I like it.

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