Find the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively.

### Concept & Logic When a number divided by divisors $a, b, c$ leaves remainders $p, q, r$ respectively, and the difference between each divisor and its remainder is a constant $K$ (i.e., $a-p = b-q = c-r = K$), the required number is the L.C.M. of the divisors minus that constant difference. $$ \text{Required Number} = \text{L.C.M.}(a, b, c) - K $$ ### Step-by-Step Solution * **Given:** Divisors are 20, 25, 35, and 40. Remainders are 14, 19, 29, and 34. * Find the common difference ($K$) between each divisor and its respective remainder: * 20 - 14 = 6 * 25 - 19 = 6 * 35 - 29 = 6 * 40 - 34 = 6 * Since the difference is constant ($K = 6$), calculate the L.C.M. of the divisors (20, 25, 35, 40). * Using prime factorization or division method, the L.C.M. of 20, 25, 35, 40 is 1400. * Subtract the common difference from the L.C.M.: * Required Number = 1400 - 6 = 1394. ### Exam Strategy & Shortcut Use **Option Elimination**. The problem states that dividing the number by 40 leaves a remainder of 34. This means if you add 6 to the correct option, it must be perfectly divisible by 40 (and 20, 25, 35). Let's test the options by adding 6: * 1394 + 6 = 1400 (Divisible by 40, ends in 00 so divisible by 25. This works perfectly). * 1400 + 6 = 1406 (Not divisible by 40). This identifies the answer in seconds without finding the LCM. ### Common Pitfall Students often intuitively *add* the constant difference (6) to the L.C.M. instead of subtracting it. Adding 6 gives 1406, which would leave a constant remainder of 6 in every case, completely violating the staggered remainders required by the question. ### Final Answer **Therefore, the correct answer is 1394.**