Question

. Find the highest number by which on dividing 118, 148 and 178, we get 2, 3
and 4 as remainders respectively.
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Answer 1

Step-by-step explanation:

Note that it is a question of HCF.

here it is being said that we have to find a highest number that will be the factor of 118 and leave 2 as remainder i.e. it will completely divide 116, similarly 148-3=145 and 178-4 =174

i.e. we require HCF of 116, 145 and 174 which is 29.

try to the HCF part by yourself.

Answer 2

Substract the respective remainders from the numbers to get the exactly divisible numbers... which come out to be 116,145 and 174.
The HCF of these 3 divide them exactly leave the aforementioned remainders

Hcf of the nos

116=29x2x2
145=29x5
174=29x3x2
Upon prime factorization the common terms arising is 29, therefore the HCF is 29