Find the least number which when divided by 15, leaves a remainder of 5, when divided by 25, leaves a remainder of 15 and when divided by 35 leaves a remainder of 25.A.515B.525C.1040D.1050

To find the least number that satisfies these conditions, we can set up a system of congruences and solve for the number using the Chinese Remainder Theorem.

Let's denote the unknown number as "x."

The given conditions can be expressed as the following congruences:

x ≡ 5 (mod 15)
x ≡ 15 (mod 25)
x ≡ 25 (mod 35)

To apply the Chinese Remainder Theorem, the moduli (15, 25, and 35) must be pairwise coprime (i.e., they have no common factors other than 1). In this case, 15, 25, and 35 are pairwise coprime, so we can proceed.

First, let's find the value of x ≡ 5 (mod 15) and x ≡ 15 (mod 25).

Using the Chinese Remainder Theorem, we can find a solution for these two congruences:

Let's find the solution for the first two congruences: x ≡ 5 (mod 15) and x ≡ 15 (mod 25).

Step 1: Solve the first pair of congruences: x ≡ 5 (mod 15) and x ≡ 15 (mod 25).

We can rewrite the first congruence as:
x ≡ 5 (mod 15)
x ≡ 0 (mod 15) + 5

For the second congruence, we can rewrite it as:
x ≡ 15 (mod 25)
x ≡ 0 (mod 25) + 15

Now we have the following system of congruences:
x ≡ 0 (mod 15) + 5
x ≡ 0 (mod 25) + 15

Step 2: Simplify the system of congruences.

We can simplify the system of congruences to the following form:
x ≡ 5 (mod 15)
x ≡ 15 (mod 25)

Step 3: Apply the Chinese Remainder Theorem.

To apply the Chinese Remainder Theorem, we need to find the modular inverses of the moduli 15 and 25.

The modular inverse of 15 (mod 25) is 8, as 15 * 8 ≡ 1 (mod 25).

Similarly, the modular inverse of 25 (mod 15) is 10, as 25 * 10 ≡ 1 (mod 15).

Step 4: Calculate the solution.

Using the Chinese Remainder Theorem, we can calculate the solution as follows:

x ≡ (5 * 25 * 8 + 15 * 15 * 10) (mod 15 * 25)
x ≡ (1000 + 2250) (mod 375)
x ≡ 3250 (mod 375)
x ≡ 125 (mod 375)

So, the solution to the first two congruences is x ≡ 125 (mod 375).

Now, let's find the value of x ≡ 125 (mod 375) and x ≡ 25 (mod 35).

Step 1: Solve the second pair of congruences: x ≡ 125 (mod 375) and x ≡ 25 (mod 35).

We can rewrite the first congruence as:
x ≡ 125 (mod 375)
x ≡ 0 (mod 375) + 125

For the second congruence, we can rewrite it as:
x ≡ 25 (mod 35)
x ≡ 0 (mod 35) + 25

Now we have the following system of congruences:
x ≡ 0 (mod 375) + 125
x ≡ 0 (mod 35) + 25

Step 2: Simplify the system of congruences.

We can simplify the system of congruences to the following form:
x ≡ 125 (mod 375)
x ≡ 25 (mod 35)

Step 3: Apply the Chinese Remainder Theorem.

To apply the Chinese Remainder Theorem, we need to find the modular inverses of the moduli 375 and 35.

The modular inverse of 375 (mod 35) is 10, as 375 * 10 ≡ 1 (mod 35).

Similarly, the modular inverse of 35 (mod 375) is 107, as 35 * 107 ≡ 1 (mod 375).

Step 4: Calculate the solution.

Using the Chinese Remainder Theorem, we can calculate the solution as follows:

x ≡ (125 * 35 * 10 + 25 * 375 * 107) (mod 375 * 35)
x ≡ (437500 + 1003125) (mod 13125)
x ≡ 1440625 (mod 13125)
x ≡ 1125 (mod 13125)

So, the solution to the last two congruences is x ≡ 1125 (mod 13125).

The final solution is x ≡ 1125 (mod 13125).

To find the least number satisfying these conditions, we can take the minimum positive solution for x.

The minimum positive solution for x is 1125.

Therefore, the least number that satisfies the given conditions is 1125.

So, the correct option is D. 1050 is not the correct answer.