Let the number =N
N=5p+2 Since the number when divided by 5 gives remainder of 2
In successive division, the quotient from first division is divided by second number.
Now p can be written as 7p+4 since p divided by 7 gives remainder of 4.
Therefore, original number can be written as:
N=5(7p+4)+2
=35p+20+2
=35p+22
If this number is divided by 35, Remainder =22

Problem: A number on being divided by 5 and 7 respectively leaves the remainder 2 and 4 respectively. Find the remainder when the same number is divided by 5×7=35.

Solution:
To find the remainder when the number is divided by 35, we need to find a number that leaves a remainder of 2 when divided by 5 and a remainder of 4 when divided by 7.

Step 1: List down the multiples of 7 and add 4 to each of them until we get a multiple of 5.
7, 14, 21, 28, 35, 42, 49, 56, 63, 70, 77, 84, 91, 98, 105, 112, 119, 126, 133, 140, 147, 154, 161, 168, 175, 182, 189, 196, 203, 210, ...

Adding 4 to each of them, we get:
11, 18, 25, 32, 39, 46, 53, 60, 67, 74, 81, 88, 95, 102, 109, 116, 123, 130, 137, 144, 151, 158, 165, 172, 179, 186, 193, 200, 207, 214, ...

Step 2: From the list above, find the smallest number that is divisible by 5.
The smallest number that is divisible by 5 is 39.

Step 3: Subtract 4 from the number obtained in Step 2 to get the required remainder.
39 - 4 = 35

Therefore, the remainder when the number is divided by 35 is 35.

Explanation:
When a number is divided by 5 and 7, the remainders obtained are 2 and 4 respectively. This means that the number can be expressed in the form:
Number = 5a + 2 = 7b + 4

We need to find the remainder when the same number is divided by 35. This means that the number can be expressed in the form:
Number = 35c + x

where x is the required remainder.

To find x, we can use the Chinese Remainder Theorem. We need to find a number that satisfies both equations:
35c + x ≡ 2 (mod 5)
35c + x ≡ 4 (mod 7)

From the first equation, we know that x ≡ 2 (mod 5). Therefore, x can be expressed as x = 2 + 5k, where k is an integer.

Substituting x = 2 + 5k in the second equation, we get:
35c + 2 + 5k ≡ 4 (mod 7)
35c + 5k ≡ 2 (mod 7)

Multiplying both sides by 5, we get:
175c + 25k ≡ 10 (mod 7)
4c + 4k ≡ 3 (mod 7)

From the above equation, we can see that c