Question

The smallest number of 4 - digits exactly divisible by 12, 15, 20 and 35 is
A. 1000
B. 1160
C. 1260
D. None

Answer 1

Hint- Proceed the solution of this question first finding the lowest common multiple (LCM) of 12, 15, 20 and 35 then divide the smallest of 4 - digits number i.e. 1000 by LCM so the remainder that we get subtract it from the sum of 1000 and LCM. Hence we get the required number.

Complete step-by-step answer:
Given number are 12, 15, 20 and 35
So their LCM, by prime factorisation
⇒12 =2×2×3

⇒15 =3×5

⇒20 =2×2×5

⇒35=5×7

So their LCM will be 2×2×3×5×7=420
We know that smallest number of 4 – digits = 1000
Hence on dividing 1000 by 420

⇒ 420 ) 1000 ( 2
            - 840
               160
We can also write it as 1000 = 420×2 + 160 (N= a×q+r) where q is quotient and r is the remainder.
So the remainder is 160
Required number = The least 4 digit number (i.e.1000) + ( Number that is exactly divisible by 12,15,20 and 35) – remainder
∴ Required number =1000+(420−160)=1260
Hence 1260 will be the smallest number of 4 - digits which is exactly divisible by 12, 15, 20 and 35.

Note- In this particular question, we should know that the Least Common Multiple (LCM) is also referred to as the Lowest Common Multiple (LCM) and Least Common Divisor (LCD). For two integers a and b, denoted LCM (a, b), the LCM is the smallest positive integer that is evenly divisible by both a and b.
As in the above question it was asked it should be lowest 4-digit number hence we further add the least 4-digit number (i.e.1000) to the Number that is exactly divisible by 12,15,20 and 35 i.e. (LCM) and later on subtracting the remainder we will get the smallest number of 4 - digits which will be exactly divisible.