Find the smallest integer greater than 100 that is divisible by 3 but leaves the
remainders 1, 5 when divided by 4 and 7 respectively.
To find the smallest integer greater than 100 that is divisible by 3 but leaves remainders 1 and 5 when divided by 4 and 7 respectively, we can follow these steps:
1. Start with the integer 101 (greater than 100).
2. Check if the number is divisible by 3 by calculating the remainder when dividing it by 3. If the remainder is not 0, increase the number by 1 and repeat step 2.
3. Check if the number leaves a remainder of 1 when divided by 4. If the remainder is not 1, increase the number by 1 and repeat step 3.
4. Check if the number leaves a remainder of 5 when divided by 7. If the remainder is not 5, increase the number by 1 and repeat step 4.
5. Once a number satisfies all conditions, it is the smallest integer greater than 100 that is divisible by 3 and leaves remainders 1 and 5 when divided by 4 and 7, respectively.
Following this process, let's find the number step-by-step:
1. Starting with the number 101.
2. 101 divided by 3 leaves a remainder of 2.
3. Increasing the number to 102.
4. 102 divided by 3 leaves a remainder of 0.
5. 102 divided by 4 leaves a remainder of 2.
6. Increasing the number to 103.
7. 103 divided by 3 leaves a remainder of 1.
8. 103 divided by 4 leaves a remainder of 3.
9. Increasing the number to 104.
10. 104 divided by 3 leaves a remainder of 2.
11. 104 divided by 4 leaves a remainder of 0.
12. 104 divided by 7 leaves a remainder of 6.
13. Increasing the number to 105.
14. 105 divided by 3 leaves a remainder of 0.
15. 105 divided by 4 leaves a remainder of 1.
16. 105 divided by 7 leaves a remainder of 0.
Therefore, the smallest integer greater than 100 that is divisible by 3 and leaves remainders of 1 and 5 when divided by 4 and 7, respectively, is 105.
To find the smallest integer greater than 100 that is divisible by 3 but leaves remainders 1 and 5 when divided by 4 and 7 respectively, you can use the concept of the Chinese Remainder Theorem.
First, let's find the number that satisfies the congruence relation for 3 and 4: x ≡ 1 (mod 4) and x ≡ 0 (mod 3).
To find a number that satisfies both congruences, we can start by finding the least common multiple (LCM) of 4 and 3, which is 12.
Now, let's list out the numbers of the form 12k + 1 (where k is a non-negative integer) that satisfy the congruence relation x ≡ 1 (mod 4):
1, 13, 25, 37, 49, ...
Out of these numbers, the smallest one that is divisible by 3 is 25.
Now, let's find the number that satisfies the congruence relation for 7 and 3: x ≡ 5 (mod 7) and x ≡ 0 (mod 3).
To find a number that satisfies both congruences, we can start by finding the LCM of 7 and 3, which is 21.
Now, let's list out the numbers of the form 21k + 5 (where k is a non-negative integer) that satisfy the congruence relation x ≡ 5 (mod 7):
5, 26, 47, 68, 89, ...
Out of these numbers, the smallest one that is divisible by 3 is 26.
Now, we need to find the number that satisfies both congruences. To do this, we can find the least common multiple of 12 and 21, which is 84.
Now, let's list out the numbers of the form 84k + 25 (where k is a non-negative integer) that satisfy both congruences:
25, 109, 193, ...
Out of these numbers, the smallest one that is greater than 100 is 109.
Therefore, the smallest integer greater than 100 that is divisible by 3 but leaves remainders 1, 5 when divided by 4 and 7 respectively is 109.
9 = 3*3 = 2*4 + 1
Now, since LCM(3,4) = 12, start adding 12's
21, 33
33 = 11*3 = 8*4 + 1 = 4*7 + 5