On dividing a number by 3, 4 and 7, the remainders are 2, 1 and 4 respectively. If the same number is divided by 84 then the remainder is
Answer: C
Let the given number be N
From the given data, we can conclude that
N = 3a + 2 = 4b + 1 = 7c + 4, where a, b and c are natural numbers.
Thus, 3a = 4b – 1 = 7c + 2
Now, 3a = 4b - 1
By trial and error, we can see that the first number satisfying the above given equation is 3
All the numbers satisfying this condition are of the form 3 + (l.c.m. of 3 and 4) × k = 3 + 12k (where k is a natural number)
Thus, 12k + 3 = 7c + 2
Hence, 12k + 1 = 7c
By trial and error method, we can see that the smallest number satisfying the above equation is 49.
Thus the smallest required number is 7c + 4 = 49 + 4 = 53
All the numbers satisfying the conditions given in the question will be of the form 53 + (l.c.m of 3, 4 and 7) × m = 53 + 84m (where m is a natural number)
∴ N when divided by 84 will result in remainder 53.
Hence, option (c).