To do that, we'll need the factor pairs of 120. (120, 1) (60, 2) (40, 3) (30, 4) (24, 5) (20, 6) (15, 8) (12, 10) Choose the pairs that have an odd and even number. Looks like 4 of them.
None that are all odd numbers. But, there are two sets of three consecutive numbers containing a mix of even/odd that add up to 120. They are 39, 40 and 41 and also 38, 40 and 42.
Yes it would Even + even = even Even + odd = odd Odd + odd = even
There are no numbers between 108 and 120 that have an even ten's digit. If you call 'zero' an even number, then ' 109 ' would qualify. But the whole exercise is still not too satisfying.
because... odd+odd=even even+odd=odd e.g 1+1+1=3 odd+odd+odd=odd
To do that, we'll need the factor pairs of 120. (120, 1) (60, 2) (40, 3) (30, 4) (24, 5) (20, 6) (15, 8) (12, 10) Choose the pairs that have an odd and even number. Looks like 4 of them.
None that are all odd numbers. But, there are two sets of three consecutive numbers containing a mix of even/odd that add up to 120. They are 39, 40 and 41 and also 38, 40 and 42.
Odd. Even + Even = Even Odd + Odd = Even Odd + Even = Even + Odd = Odd
odd * odd = odd answer even * even = even answer odd * even = even answer
even times even = even odd times even = even odd times odd = odd
Yes. Even + Even = Even, Odd + Odd = Even and Even + Odd or Odd + Even = Odd
No. Sum of odd + odd = even Sum of odd + even = odd Sum of even + even = even
To find the largest odd natural number that is a factor of 120, we first need to factorize 120 into its prime factors, which are 2^3 * 3 * 5. Since we are looking for an odd factor, we can ignore the factor of 2. The largest odd factor of 120 is then 3, as it is the largest odd prime factor present in the prime factorization of 120.
Always. even + even = even odd + odd = even even + odd = odd odd + even = odd To summarise, if you add like numbers you get even, otherwise you get odd.
even times even = even odd times even = even odd times odd = odd
it is even
even times even = even odd times odd = odd even times odd = even