Question 72

What is the highest number which when divides the numbers 1026, 2052 and 4102, leave remainders 2, 4 and 6 respectively?

Solution

Let's assume the highest number is 'y' when divides the numbers 1026, 2052 and 4102, leave remainders 2, 4 and 6 respectively.

So 1026-2 = 1024

2052-4 = 2048

4102-6 = 4096

Now we need to take the HCF of 1024, 2048 and 4096.

1024 = 1024

2048 = $$1024\times2$$

4096 = $$1024\times4$$

So the HCF for these = y = 1024

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