Find the greatest number that divides 273 and 312 exactly without leaving any remainder
Answers
Answer:
Answer:
GCF = 39
Step by Step explanation:
For the values 273, 312
Solution by Factorization:
The factors of 273 are: 1, 3, 7, 13, 21, 39, 91, 273
The factors of 312 are: 1, 2, 3, 4, 6, 8, 12, 13, 24, 26, 39, 52, 78, 104, 156, 312
Then the greatest common factor is 39.
Euclid's Algorithm
What do you do if you want to find the GCF of more than two very large numbers such as 182664, 154875 and 137688? It's easy if you have a Factoring Calculator or a Prime Factorization Calculator or even the GCF calculator shown above. But if you need to do the factorization by hand it will be a lot of work.
How to Find the GCF Using Euclid's Algorithm
Given two whole numbers, subtract the smaller number from the larger number and note the result.
Repeat the process subtracting the smaller number from the result until the result is smaller than the original small number.
Use the original small number as the new larger number. Subtract the result from Step 2 from the new larger number.
Repeat the process for every new larger number and smaller number until you reach zero.
When you reach zero, go back one calculation: the GCF is the number you found just before the zero result.
Example: Find the GCF (18, 27)
27 - 18 = 9
18 - 9 - 9 = 0
So, the greatest common factor of 18 and 27 is 9, the smallest result we had before we reached 0.
Example: Find the GCF (20, 50, 120)
Note that the GCF (x,y,z) = GCF (GCF (x,y),z). In other words, the GCF of 3 or more numbers can be found by finding the GCF of 2 numbers and using the result along with the next number to find the GCF and so on.
Let's get the GCF (120,50) first
120 - 50 - 50 = 120 - (50 * 2) = 20
50 - 20 - 20 = 50 - (20 * 2) = 10
20 - 10 - 10 = 20 - (10 * 2) = 0
So, the greatest common factor of 120 and 50 is 10.
Now let's find the GCF of our third value, 20, and our result, 10. GCF (20,10)
20 - 10 - 10 = 20 - (10 * 2) = 0
So, the greatest common factor of 20 and 10 is 10.
Therefore, the greatest common factor of 120, 50 and 20 is 10.
Example: Find the GCF (182664, 154875, 137688) or GCF (GCF(182664, 154875), 137688)
First we find the GCF (182664, 154875)
182664 - (154875 * 1) = 27789
154875 - (27789 * 5) = 15930
27789 - (15930 * 1) = 11859
15930 - (11859 * 1) = 4071
11859 - (4071 * 2) = 3717
4071 - (3717 * 1) = 354
3717 - (354 * 10) = 177
354 - (177 * 2) = 0
So, the the greatest common factor of 182664 and 154875 is 177.
Now we find the GCF (177, 137688)
137688 - (177 * 777) = 159
177 - (159 * 1) = 18
159 - (18 * 8) = 15
18 - (15 * 1) = 3
15 - (3 * 5) = 0
So, the greatest common factor of 177 and 137688 is 3.
Therefore, the greatest common factor of 182664, 154875 and 137688 is 3.
Solution:
Solution:Set up a division problem where a is larger than b.
Solution:Set up a division problem where a is larger than b.a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.
Solution:Set up a division problem where a is larger than b.a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.312 ÷ 273 = 1 R 39 (312 = 1 × 273 + 39)
Solution:Set up a division problem where a is larger than b.a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.312 ÷ 273 = 1 R 39 (312 = 1 × 273 + 39)273 ÷ 39 = 7 R 0 (273 = 7 × 39 + 0)
Solution:Set up a division problem where a is larger than b.a ÷ b = c with remainder R. Do the division. Then replace a with b, replace b with R and repeat the division. Continue the process until R = 0.312 ÷ 273 = 1 R 39 (312 = 1 × 273 + 39)273 ÷ 39 = 7 R 0 (273 = 7 × 39 + 0)When remainder R = 0, the GCF is the divisor, b, in the last equation. GCF = 39