Find the largest number which divides 64, 136 and 238 to leave the same remainder in each case.
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Answered by
6
Let a be one of the factors of 62 132 237 and c be the reminder when divisible by a (which is common)
ab1 + c =62
ab2 + c = 132
ab3 + c = 237
Solving three equations
a(b2-b1) = 70
a(b3-b1) = 175
a(b3-b2)=105
There are multiple combinations of a. to find the largest of them, we need to find the HCF of 70, 105, 175. which is 35.
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ab1 + c =62
ab2 + c = 132
ab3 + c = 237
Solving three equations
a(b2-b1) = 70
a(b3-b1) = 175
a(b3-b2)=105
There are multiple combinations of a. to find the largest of them, we need to find the HCF of 70, 105, 175. which is 35.
hope it's helpful if yes then mark this answer as brainliest ......
and if you don't know how to make this answer is brainliest then message me .....
Thank you
Answered by
0
Answer:
Step-by-step explanation:
6 is the correct answer
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