find the largest positive integer that will divide 122,150 and 115 leaving remainder 5 , and 11 respectively
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Let that largest positive integer be x.
Given, when 122 is divided by x, it leaves the remainder 5
Thus, (122-5) is divisible by x, i.e. 117 is divisible by x.
Given, when 150 is divided by x, it leaves remainder 7
Thus, (150-7) is divisible by x, i.e. 143 is divisible by x.
Given, when 115 is divided by x, it leaves remainder 11
Thus, (115-11) is divisible by x, i.e. 104 is divisible by x.
Hence, x = H. C. F. of 117, 143, 104.
Factors of 117 = 3 × 3 × 13
Factors of 143 = 11 × 13
Factors of 104 = 2 × 2 × 2 × 13
Thus, H. C. F. of 117, 143, 104 = 13
Then, 13 is the largest number which will divide 117, 143, 104 and leaving the remainders as 5, 7, 11.
Step-by-step explanation:
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