Answer:

As the largest positive integer that will divide 122, 150 and 115 leaves remainder 5, 7 and 11 respectively.

Therefore,

122−5=117

150−7=143

115−11=104

∵ The HCF of 104, 117 and 143 is 13

Hence, the largest number which divides 122, 150 and 115  leaving 5, 7 and 11 as remainders respectively is 13