There is more than one integer greater than 1 which leaves a reminder of1 when divided by each of the four smallest primes

210

Complete question found at (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

1. supplementary angles

2. acd and bcd

step-by-step explanation:

in total they made   333.74 your welcome : d