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Find the greatest number which divides 38, 47 and 76 leaving remainder 3, 2 and 1 respectively.
Given: 38, 47 and 76
To find: Here we have to find a number that divides 38, 47 and 76 leaving the remainder 3, 2 and 1 respectively.
Solution:
If we subtract the remainders, then we will get the numbers that are exactly divisible by the unknown number.
38 $-$ 3 = 35
47 $-$ 2 = 45
76 $-$ 1 = 75
The answer will be HCF OF 35, 45 and 75:
75 = 3 $ \times $ 52
35 = 5 $ \times $ 7
45 = 32 $ \times $ 5
The highest common factor of 75, 35 and 45 is 5.
So, the number that divides 38, 47 and 76 leaving the remainder 3, 2 and 1 respectively is 5.
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