The greatest number which divides 281 and 1249 , leaving remainder 5 and 7 respectively, is :\

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Professor PC Thomas Classes · Answer

(c): We have to find the greatest number which divides 281 and 1249, leaving remainder 5 and 7 respectively. This means that $281-5=276$ and $1249-7=1242$ are divisible by the required number.\ $	herefore \quad$ Required number is HCF of 276 and 1242.\ Prime factorisation of 276 and 1242 are $$ 276=2 	imes 2 	imes 3 	imes 23 $$ $1242=2 	imes 3 	imes 3 	imes 3 	imes 23$\ $	herefore \quad \operatorname{HCF}(276,1242)=2 	imes 3 	imes 23=138$\