The greatest number, which divides 134 and 167 leaving 2 as remainder in each
case, is
(a) 11
(b) 17
(c) 13
(d) 12
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The largest number which divides 615 and 963 leaving remainder 2 in each case => HCF.
And let the HCF be x.
To make 134 and 167 completely divisible by x we have to sub. 2 from 134 and 167.
134-2=132,167-2=165
165 = 3*5*11
132= 2*2*3*11
⇒ x =3*11⇒x=33
Therefore the greatest number which divides 134 and 167 leaving 2 as remainder in each case as 33.
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