Find the smallest and largest three-digit numbers which when divided by 22,33 and 55 leave remainder of 5 in each case.
(a). 340, 980
(b). 335, 995
(c). 330, 990
(d). 325, 985

Hint: In this problem, we will find the LCM (lowest common multiple) of the three numbers 22, 33 and 55. We can do this by listing prime factors and appropriately combining them to get the LCM. Then we will find the largest and smallest multiple of this LCM (which is a 3-digit number) and then add 5 to this number to find the required answer.

Complete step-by-step answer:

First, we will briefly understand prime factorisation. Prime factorization is a process of factoring a number in terms of prime numbers i.e. the factors will be prime numbers. The first few prime numbers are 2, 3, 5, 7, 11, 13, 17, 19 and so on. These prime numbers when multiplied with any natural numbers produce composite numbers. For example, Prime factorization of 12 is 2 x 2 x 3 = 22 x 3
Let us now consider the given question,

Prime factorisation of 22=2×11
Prime factorisation of 33=3×11
Prime factorisation of 55=5×11
To find LCM (Least Common Multiple) we only take the product of distinct factors of these 3 numbers. Thus, we have,
LCM 22,33,55=2×3×5×11=330


As 330 is the smallest 3 digit number divisible by 22,33,55, the number 330+5=335 will give a remainder 5 when divided by these numbers.
The highest 3 digit number is 999.

999 when divided by 330 gives a remainder 9, so 999−9=990 is the largest 3− digit number divisible by 22,33,55, the number 990+5=995 will give a remainder 5 when divided by these numbers.

Therefore, the final answer is (b) 335, 995.

Note: In questions which involve finding a number which can divide all n (here, n=3) numbers. We do this by finding the LCM (Least Common Multiple) of the three numbers. This gives the smallest number that can divide all the numbers. One can then add the relevant constraints depending on the question to get the required answer.