The least number which on division by 35 leaves a remainder 25 and on division by 45 leaves the remainder 35 and on division by 55 leaves the remainder 45 is:(a). 2515(b). 3455(c). 2875(d). 2785

We are asked to find the least number that satisfies the following conditions:

When divided by 35, the remainder is 25.
When divided by 45, the remainder is 35.
When divided by 55, the remainder is 45.
We can express these conditions using modular arithmetic:

Let the number be denoted by x.

x ≡ 25 (mod 35)
x ≡ 35 (mod 45)
x ≡ 45 (mod 55)
We notice that the remainders are always 10 less than the divisors. This suggests that if we add 10 to the number x, we should have the following system of congruences:

x + 10 ≡ 0 (mod 35)
x + 10 ≡ 0 (mod 45)
x + 10 ≡ 0 (mod 55)
Now, the number x + 10 must be a multiple of the least common multiple (LCM) of 35, 45, and 55.

Step 1: Find the LCM of 35, 45, and 55
We first find the prime factorizations of these numbers:

35 = 5 × 7
45 = 3² × 5
55 = 5 × 11
The LCM is found by taking the highest powers of all prime factors present:

LCM = 3² × 5 × 7 × 11 = 3² × 5 × 7 × 11 = 9 × 5 × 7 × 11 = 3465
Thus, x + 10 must be a multiple of 3465.

Step 2: Find the least value of x
The smallest possible value for x + 10 is 3465, so:

x + 10 = 3465 x = 3465 - 10 = 3455

Step 3: Verify the solution
We now check if x = 3455 satisfies the original conditions:

3455 ÷ 35 = 98 remainder 25. (Condition satisfied)
3455 ÷ 45 = 76 remainder 35. (Condition satisfied)
3455 ÷ 55 = 62 remainder 45. (Condition satisfied)
Thus, the least number that satisfies all the given conditions is 3455.

The answer is (b) 3455.