Question

The smallest number which when divided by 20, 25, 35, and 40 leaves remainders 14, 19, 29, and 34 is
(a) 1394
(b) 1404
(c) 1664
(d) 1406

Answer 1

Hint: Here, we will write the given divisors as a product of their prime factors. Then, we will calculate the L.C.M. of the four divisors. Using the given remainders, we will find the smallest number to be deducted from the L.C.M. to get the required number which when divided by 20, 25, 35, and 40 leaves remains 14, 19, 29, and 34. The lowest common multiple is the product of the prime factors with the greatest powers.

Complete step-by-step answer:
The given divisors are 20, 25, 35, and 40.
We will find the L.C.M. of the divisors using the fundamental theorem of arithmetic.
First, we will write the given numbers as a product of their prime factors.
We know that 20 is the product of 4 and 5.
Therefore, we can write 20 as
\[20 = 4 \times 5\]
4 is the square of the prime number 2. Thus, we get
\[ \Rightarrow 20 = {2^2} \times 5\]
Now, we know that 25 is the square of 5.
Therefore, we can write 25 as
\[25 = {5^2}\]
35 is the product of the prime numbers 5 and 7. Thus, we get
\[35 = 5 \times 7\]
We know that 40 is the product of 8 and 5.
Therefore, we can write 40 as
\[40 = 8 \times 5\]
8 is the cube of the prime number 2. Thus, we get
\[ \Rightarrow 40 = {2^3} \times 5\]
Therefore, we have
\[20 = {2^2} \times 5\]
\[25 = {5^2}\]
\[35 = 5 \times 7\]
\[40 = {2^3} \times 5\]
Now, in the product of primes, we can observe that the greatest power of 2 is 3, greatest power of 5 is 2, and the greatest power of 7 is 1.
Thus, the prime factors with the greatest powers are \[{2^3}\], \[{5^2}\], and 7.
The lowest common multiple of the numbers 20, 25, 35, 40 is the product of the prime factors with the greatest powers.
Therefore, we get
\[L.C.M. = {2^3} \times {5^2} \times 7\]
Simplifying the expression, we get
\[\begin{array}{c} \Rightarrow L.C.M. = 8 \times 25 \times 7\\ \Rightarrow L.C.M. = 200 \times 7\end{array}\]
Multiplying the terms, we get
\[ \Rightarrow L.C.M. = 1400\]
\[\therefore \] The L.C.M. of the divisors is 1400.
The required number when divided by 20, 25, 35, 40 leaves 14, 19, 29, 34 as remainder respectively.
We can observe that
\[\begin{array}{l}20 - 14 = 6\\25 - 19 = 6\\35 - 29 = 6\\40 - 34 = 6\end{array}\]
The difference of the divisors and the respective remainders is always 6.
Therefore, the required smallest number which when divided by 20, 25, 35, and 40 leaves remainders 14, 19, 29, and 34 is the difference in the L.C.M. of the divisors, and the number 6.
Thus, we get
Required number \[ = 1400 - 6\]
Subtracting the terms in the expression, we get
\[ \Rightarrow \] Required number \[ = 1394\]
Therefore, the smallest number which when divided by 20, 25, 35, and 40 leaves remains 14, 19, 29, and 34 is 1394.
Thus, the correct option is option (a).

Note: All the prime factors with the greatest powers are selected, irrespective of whether that power appears in the prime factorization of the divisors. For example, \[{2^3}\] does not appear in the prime factorization of 20, 25, and 35. But it should be included while calculating L.C.M. because it has the highest power. Another common mistake is to use the common factors with the lowest powers to calculate the L.C.M. That is incorrect because it will give you the H.C.F. and not the L.C.M. of the numbers.
We used the fundamental theorem of arithmetic in the solution. The fundamental theorem of arithmetic states that every composite number can be written as a product of its prime factors in a unique way.