Question

Let x be a least number which when divided by 21, 33,35 and 55 leaves a remainder of 3. But is exactly divisible by 67

Answer 1

Suppose  we are given that a number when divided by x, y, and z, leaves a  remainder of a, b, and c; then the number will be of the format of 
LCM(x,y,z)*n + constant

The  key in these questions is finding out the value of 'constant'. If all  of them leave the same remainder 'r', constant = r. It can also be  looked at as the smallest number satisfying the given property.

In this question, we are given 
Remainder from 18 is 2
Remainder from 35 is 19
Remainder from 42 is 26

If you look at the negative remainders
Remainder from 18 is -16
Remainder from 35 is -16
Remainder from 42 is -16

So, the number N = LCM(18,35,42)*n - 16 = 630n - 16
So, any number which is of the format of 630n - 16 will satisfy the given conditions.
Least possible number would occur when n = 1
Least possible number will be 630 - 16 = 614

Answer 2

Answer:

15

Step-by-step explanation:

A/Q

x= 21 k1  +3

x =33 k2 +3

x =35 k3 +3

x =55 k4 +3

x  =67 k5

= LCM (211 331 351 55)k + 3

= 1155k + 3

A/Q x = 67 k5

67 k5= 1155 k + 3  

at k = 4

Number = 4623 which

satisfies all conditions.

sum of digits = 4 + 6 + 2 + 3 = 15