ok this will be very easy as i got it in just a few seconds myself but its fun to do nonetheless. find a number that is:
a number between 1 & 36
two digits
a multiple of 3
when the digits are added together the sum is between 4 & 8
is odd
when you multiply the digits the product is between 4 & 8
what is the number?
Originally posted by tournymangr15 r 17
ok this will be very easy as i got it in just a few seconds myself but its fun to do nonetheless. find a number that is:
a number between 1 & 36
two digits
a multiple of 3
when the digits are added together the sum is between 4 & 8
is odd
when you multiply the digits the product is between 4 & 8
what is the number?
Originally posted by joe shmoYup, it's a way to check if a number is a multiple of three;
Has anyone noticed the sum of the digits of multiples of 3 repeat 3,6,9 up until 39 which its sum is 12, which has a sum of 3 restarting the sequense replacing a sum of 12 fo three. can anyone explain this?
Have a number abc (where a, b and c stand for the single digits of a three-digit number).
You can write it as a * 100 + b * 10 + c * 1
a * 99 is a multiple of 3, as well as b * 9.
So abc is a multiple of three
if and only if
a * 1 + b * 1 + c * 1 = a + b + c is a multiple of three.
This works for all numbers.
So to check if 123456723 is a multiple of three, we calculate 1+2+3+4+5+6+7+2+3 = 33.
To check if 33 is a multiple of three, we calculate 3+3 = 6
6 is a multiple of 3, so 123456723 is a multiple of three as well.
You have a similar trick for multiples of 9 (sum of digits must be 9 then) and multiples of 11 (alternating sum of digits must be 0).
There is even one for multiples of 7, but that's a bit trickier.
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Originally posted by TheMaster37Thanks, I find that interesting.
Yup, it's a way to check if a number is a multiple of three;
Have a number abc (where a, b and c stand for the single digits of a three-digit number).
You can write it as a * 100 + b * 10 + c * 1
a * 99 is a multiple of 3, as well as b * 9.
So abc is a multiple of three
if and only if
a * 1 + b * 1 + c * 1 = a + b + c is a multiple of ...[text shortened]... sum of digits must be 0).
There is even one for multiples of 7, but that's a bit trickier.
Originally posted by TheMaster37or, you can just say that if the sum of digits is divisible by three, then the number is divisible by three...
Yup, it's a way to check if a number is a multiple of three;
Have a number abc (where a, b and c stand for the single digits of a three-digit number).
You can write it as a * 100 + b * 10 + c * 1
a * 99 is a multiple of 3, as well as b * 9.
So abc is a multiple of three
if and only if
a * 1 + b * 1 + c * 1 = a + b + c is a multiple of ...[text shortened]... sum of digits must be 0).
There is even one for multiples of 7, but that's a bit trickier.
you can do this multiple times if necessary...
for example:
111111111111111111111111111111111111111111111111111111111
sum of digits is 57 you can say that 57 is divisible by 3 since 5+7 = 12 and 1+2 = 3...
meanin that the entire number is divisible by three.