What is the smallest positive integer that leaves a remainder of 1 when divided by 12, 25 and 2020
Answers
Step-by-step explanation:
If and are positive integers, there exist unique integers and , called the quotient and remainder, respectively, such that and .
For example, when 15 is divided by 6, the quotient is 2 and the remainder is 3 since .
Notice that means that remainder is a non-negative integer and always less than divisor.
This formula can also be written as .
Answer:
The smallest positive integer that leaves a remainder of 1 when divided by 12, 25, and 2020 = 30301
Step-by-step explanation:
Given numbers are 15,25,2020
To find,
The smallest positive integer that leaves remainder 1 when divided by 12,25, and 2020
LCM is the smallest positive number that leaves the remainder '0'
So, the smallest positive number which leaves the remainder '1' is LCM +1
Hence, we need to find LCM(12,25,2020) +1
12 = 2 × 2 × 3
25 = 5 x 5
2020 = 2 × 2 × 5 × 101
Hence LCM(12,25,2020) = 2 ×2 × 3 × 5 × 5 × 101 = 30300
LCM (12,25,2020) = 30300
The smallest positive integer that leaves a remainder of 1 when divided by 12, 25 and 2020 = LCM (12,25,2020) +1
= 30300+1 = 30301
∴The smallest positive integer that leaves a remainder of 1 when divided by 12, 25, and 2020 = 30301
#SPJ3