 # RD Sharma Solutions Class 10 Mathematics Solutions for Real Numbers Exercise 1.2 in Chapter 1 - Real Numbers

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

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 12625 to be 625.

∴ H.C.F. (625, 3125, 15625) = 625

So, the required number is 625.

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