find largest number that divides 62. 142 .236 leaves same remainder in each case
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The given numbers are 62, 142 and 236
They leave same remainder when divided by a certain number. Let it be x.
If x divides 62 and 142 and leaves same remainder, then it will divided their difference leaving no remainder. (Try if you are not convinced)
142 - 62 = 80
236 - 142 = 94
236 - 62 = 174
So x will divide 80, 94 and 174 leaving no remainder.
To find the greatest possible value of x, find HCF of 80, 94 and 174
80 = 2*2*2*2*5
94 = 2*47
236 = 2*2*59
HCF is 2.
So the largest number which divides 62,142 and 236 leaving the same remainder in each case is 2.
They leave same remainder when divided by a certain number. Let it be x.
If x divides 62 and 142 and leaves same remainder, then it will divided their difference leaving no remainder. (Try if you are not convinced)
142 - 62 = 80
236 - 142 = 94
236 - 62 = 174
So x will divide 80, 94 and 174 leaving no remainder.
To find the greatest possible value of x, find HCF of 80, 94 and 174
80 = 2*2*2*2*5
94 = 2*47
236 = 2*2*59
HCF is 2.
So the largest number which divides 62,142 and 236 leaving the same remainder in each case is 2.
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