1, 2, 4, 8, 16.
1, 2, 4, 8, 16, 32.
Each of these numbers will go evenly into 80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.
These whole numbers go evenly into 64: 1, 2, 4, 8, 16, 32, 64.
20% of 80 is 16 and 80-16= 64
The largest number that will divide evenly into 64 with no remainder is 64.
The HCF of 16 and 64 is 16
no but 2 goes evenly into 62 and 80 just do 31x2=62 and 40x=80
1, 2, 4, 8, 16, 32, 64
Yes, it goes in 16 times (16x4=64)
27 is 3 to the 3rd power, and 64 is 2 to the 6th power, so the only number that goes into both evenly is one (1).
1, 2, 4, 8, 16, 32, 64.
The only common factor of 49 and 64 (whole number that goes into both of them evenly) is ' 1 '.
1, 2, 4, 8, 16, 32.
No because it will have a remainder of 12
Each of these numbers will go evenly into 80: 1, 2, 4, 5, 8, 10, 16, 20, 40, 80.
2 2 goes into any even number right away evenly. In this case though, the highest number would be 4.
Sixteen is a divisor of 64 because it goes evenly into 64. Here are all of 64's divisors: 1, 2, 4, 8, 16, 32, 64.