To find the least number which, when divided by 15, 25, 35, and 40, leaves remainders of 10, 20, 30, and 35 respectively, we can follow these steps:
### Step 1: Understand the problem
We need to find a number \( N \) such that:
- \( N \mod 15 = 10 \)
- \( N \mod 25 = 20 \)
- \( N \mod 35 = 30 \)
- \( N \mod 40 = 35 \)
### Step 2: Convert the conditions
From the conditions given, we can rewrite them as:
- \( N = 15k + 10 \) for some integer \( k \)
- \( N = 25m + 20 \) for some integer \( m \)
- \( N = 35n + 30 \) for some integer \( n \)
- \( N = 40p + 35 \) for some integer \( p \)
### Step 3: Rearranging the equations
We can rearrange these equations to find a common form:
- From \( N \mod 15 = 10 \), we can say \( N - 10 \equiv 0 \mod 15 \) or \( N \equiv 10 \mod 15 \).
- From \( N \mod 25 = 20 \), we can say \( N - 20 \equiv 0 \mod 25 \) or \( N \equiv 20 \mod 25 \).
- From \( N \mod 35 = 30 \), we can say \( N - 30 \equiv 0 \mod 35 \) or \( N \equiv 30 \mod 35 \).
- From \( N \mod 40 = 35 \), we can say \( N - 35 \equiv 0 \mod 40 \) or \( N \equiv 35 \mod 40 \).
### Step 4: Finding the least common multiple (LCM)
To solve these congruences, we first need to find the least common multiple (LCM) of the divisors:
- The numbers are 15, 25, 35, and 40.
Calculating the LCM:
- Prime factorization:
- \( 15 = 3 \times 5 \)
- \( 25 = 5^2 \)
- \( 35 = 5 \times 7 \)
- \( 40 = 2^3 \times 5 \)
Taking the highest power of each prime:
- \( 2^3 \) from 40
- \( 3^1 \) from 15
- \( 5^2 \) from 25
- \( 7^1 \) from 35
Thus, the LCM is:
\[
LCM = 2^3 \times 3^1 \times 5^2 \times 7^1 = 8 \times 3 \times 25 \times 7
\]
Calculating this step-by-step:
- \( 8 \times 3 = 24 \)
- \( 24 \times 25 = 600 \)
- \( 600 \times 7 = 4200 \)
So, the LCM of 15, 25, 35, and 40 is 4200.
### Step 5: Finding the least number
Now, since \( N \equiv 10 \mod 15 \), \( N \equiv 20 \mod 25 \), \( N \equiv 30 \mod 35 \), and \( N \equiv 35 \mod 40 \), we need to subtract the common remainder (which is 5) from the LCM:
\[
N = 4200 - 5 = 4195
\]
### Final Answer
The least number which, when divided by 15, 25, 35, and 40, leaves remainders of 10, 20, 30, and 35 respectively is **4195**.
