Answer:
210
Complete question found at brainly(ID): 18678557 is stated below.
There is more than one integer, greater than 1, which leaves a remainder of 1 when divided by each of the four smallest primes. What is the difference between the two smallest such integers?
Step-by-step explanation:
Prime numbers are numbers that can only be divided by itself and 1
The smallest of the prime numbers we have = 2, 3, 5, 7
Since the integers greater than 1 are said to be divided by the four smallest prime numbers, we would assume the number of integers are 4 in total.
Let the integers be T
From the question:
Integer/(prime number) = quotient + (remainder/prime number)
Integer/(prime number) = Q + R/P
Let the different quotients derived from all 4 prime number = w, x, y, z
For prime 2:
T/2 = w + 1/2
T/2 - 1/2 = w
(T-1)/2 = w
T = 2w + 1
T-1 = 2w
Following the above solution
For prime 3:
T = 3x + 1
T-1 = 3x
For prime 5:
T = 5y + 1
T-1 = 5y
For prime 7:
T = 7z + 1
T-1 = 7z
T-1= T-1 = T-1 = T-1
2w = 3x = 5y = 7z
T-1 = LCM of all the prime numbers
T- 1 = 2×3×5×7
T-1 = 210
T = 210+1 = 211
T = 211
The smallest of the integer greater than 1 that leaves a remainder of 1 = 1(T-1) + 1 = 211
The next after the smallest number: 2(T-1) +1= 2(210) + 1 = 421
The two smallest number = 1(T-1) + 1 and 2(T-1) +1 respectively
The difference between the two smallest such integers = 421-211 = 210
