This is the basic and exam oriented question from real numbers in which we have given two numbers 438 and 606 and we have to find the largest number which divides the above number leaving remainder 6 in each case.
Kindly solve the above problem
RS Aggarwal, Class 10, chapter 1B, question no 8
Largest number which divides 438 and 606, leaving remainder 6 is actually the largest number which divides 438 – 6 = 432 and 606 – 6 = 600, leaving remainder 0. Therefore, HCF of 432 and 600 gives the largest number.
Now, prime factors of 432 and 600 are:
432 = 2^4 × 3³
600 = 2³ × 3 × 5²
HCF = product of smallest power of each common prime factor in the numbers = 2³ × 3 = 24
Thus, the largest number which divides 438 and 606, leaving remainder 6 is 24.