Question

The least possible number of 3 digits when successively divided by 2, 5, 4, 3 gives respective remainders of 1, 1, 3, 1 is:

• A. 372
• B. 275
• C. 273
• D. 193

Answer 1

The correct option is D 193

The problem can be expressed as

So it can be solved as
(((((E $\times$ 3) + 1) $\times$ 4 + 3) 5 + 1 ) 2 + 1) = A
(where A is the required number)
so for the least possible number E = 1 (the least positive integer)
then, A = (((((E $\times$ 3) + 1) $\times$ 4 + 3) 5 + 1 ) 2 + 1)
[Since at E = 0, we get a two digit number]
Alternatively: We use the following convention while solving this type of problem. First we write all the divisors as given below then their respective remainders just below them.

Where the arrow downwards means to add up and the arrow slightly upward (at an angle of ${45}^{\circ }$) means multiplication.
So we start from the right side remainder and move towards left since while writing down the divisors and remainders we write the first divisor first (i.e., leftmost), second divisor at second position (from the left) and so on.
Now solve it as: Step 1. (1 $\times$ 4) + 3 = 7
Step 2. 7 $\times$ 5 + 1 = 36
Step 3. 36 $\times$ 2 + 1 = 73
or ((((1 $\times$ 4) + 3) 5 + 1) 2 + 1) = 73
but we are required to find a three digit numbr so the next higher numbers can be obtained just by taking the multiples of the product of the divisors and then adding to it the least such number.
The next higher number $=(2\times 5\times 4\times 3)$m + 73 = 120m + 73
So by putting m = 1, 2, 3, . . . we can get the higher possible numbers.
Now we just need a least possible 3 digit number so we can get it by putting m = 1
Hence the required number = 120 $\times$ 1 + 73 = 193
Hence (d) is the correct answer.