Question #18c65

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Answer 1 · 2017-08-22T18:08:41

#GCD=7#

With larger numbers using the Euclidean algorithm is helpful

#GCD(595,252)#

#595=252xx2+91#

#252=91xx2+70#

#91=70xx1+21#

#70=21xx3+7#

#21=3xx7+0#

The last non zero remainder is the #GCD#

#GCD=7#

Answer 2 · 2017-08-25T18:40:22

They are both divisible by #color(red)(7)#

Write each number as the product of its prime factors:

#595 = color(white)(xxxxxxxxxxx)5 xx color(red)(7) xx 17#
#252 =2xx2xx3xx3xx" "color(red)(7)#

The only factor that is common is #color(red)(7)# so that will be the highest divisor.