Question

What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?

Answer 1

The  largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively is 625.

Given,

When 626 is divided by a number, it leaves a remainder 1

⇒ 626 - 1 = 625

The factors of 625 = 5 × 5 × 5 × 5

When 3127 is divided by a number, it leaves a remainder 2

⇒ 3127 - 2 = 3125

The factors of 3125 = 5 × 5 × 5 × 5 × 5

When 15628 is divided by a number, it leaves a remainder 3

⇒ 15628 - 3 = 15625

The factors of 15625 = 5 × 5 × 5 × 5 × 5 × 5

Therefore, the HCF of all the above numbers is 5 × 5 × 5 × 5 = 625

Hence, 625 is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.

Answer 2

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From the question it’s understood that,

626 – 1 = 625, 3127 – 2 = 3125 and 15628 – 3 = 15625 has to be exactly divisible by the

number.

Thus, the required number should be the H.C.F of 625, 3125 and 15625.

First, consider 625 and 3125 and apply Euclid’s division lemma

3125 = 625 x 5 + 0

∴ H.C.F (625, 3125) = 625

Next, consider 625 and the third number 15625 to apply Euclid’s division lemma

15625 = 625 x 25 + 0

We get, the HCF of 625 and 15625 to be 625.

∴ H.C.F. (625, 3125, 15625) = 625

So, the required number is 625.