The least number which when divided by 35 leaves remainder of 25 when divided with 45 leaves a remainder of 35 and when divided by 55 leaves 45 as remainder is?

Answer 1

To find the least number that meets these conditions, we can express the problem in terms of congruences. We have:

1. ( x \equiv 25 \mod 35 )
2. ( x \equiv 35 \mod 45 )
3. ( x \equiv 45 \mod 55 )

These congruences can be rewritten as:

• ( x \equiv -10 \mod 35 )
• ( x \equiv -10 \mod 45 )
• ( x \equiv -10 \mod 55 )

Since all three congruences have the same form, we can solve for ( x ) using the least common multiple of the moduli (35, 45, and 55), which is 1035. Adding 10 to account for the negative remainder, we find that ( x = 1035 - 10 = 1025 ).

Thus, the least number is 1025.