Asked by Anonymous
find the largest number that will divide 220,313,and 716 leaving remainder 3 in each case?
Answers
Answered by
Steve
217 = 7*31
310 = 10*31
713 = 23*31
It appears that dividing any of the original numbers by 31 will leave a remainder of 3.
310 = 10*31
713 = 23*31
It appears that dividing any of the original numbers by 31 will leave a remainder of 3.
Answered by
Bosnian
220 - 3 = 217
313 - 3 = 310
716 - 3 = 713
Find the prime factorization of each integer.
217 = 7 ∙ 31
310 = 2 ∙ 5 ∙ 31
713 = 23 ∙ 31
The "Greatest Common Factor " is the largest of the common factors (of two or more numbers)
In this case GCF = 31
Because 31 is the greatest number that divides evenly into all of them.
The largest number that will divide 220,313,and 716 leaving remainder 3 = 31
220 / 31 = ( 217 + 3 ) / 31 = 217 / 31 + 3 / 31 = 7 + 3 / 31
A remainder = 3
313 / 31 = ( 310 + 3 ) / 31 = 310 / 31 + 3 / 31 = 10 + 3 / 31
A remainder = 3
716 / 31 = ( 713 + 3 ) / 31 = 713 / 31 + 3 / 31 = 23 + 3 / 31
A remainder = 3
313 - 3 = 310
716 - 3 = 713
Find the prime factorization of each integer.
217 = 7 ∙ 31
310 = 2 ∙ 5 ∙ 31
713 = 23 ∙ 31
The "Greatest Common Factor " is the largest of the common factors (of two or more numbers)
In this case GCF = 31
Because 31 is the greatest number that divides evenly into all of them.
The largest number that will divide 220,313,and 716 leaving remainder 3 = 31
220 / 31 = ( 217 + 3 ) / 31 = 217 / 31 + 3 / 31 = 7 + 3 / 31
A remainder = 3
313 / 31 = ( 310 + 3 ) / 31 = 310 / 31 + 3 / 31 = 10 + 3 / 31
A remainder = 3
716 / 31 = ( 713 + 3 ) / 31 = 713 / 31 + 3 / 31 = 23 + 3 / 31
A remainder = 3
Answered by
Anonymous
31
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