When the HCF of two numbers is found by the long division (continued division)…

Concept

In the continued-division (Euclidean) method for HCF, each step feeds the next: the divisor of one step becomes the dividend of the next, and the remainder of one step becomes the divisor of the next. The process stops when the remainder is 0, and the last divisor is the HCF.

So if we know the last divisor (the HCF) and every quotient, we can rebuild the original two numbers backwards using the division identity:

Dividend = (Divisor × Quotient) + Remainder.

Applying it here

Last divisor (HCF) = 64; quotients from first to last = 2, 7, 2. Work the steps from the last one upward:

1. Last step — the remainder is 0 here (the division terminates). Divisor = 64, quotient = 2 (the last quotient).

   Dividend = (64 × 2) + 0 = 128.

2. Middle step — its divisor was the previous dividend (128), and its remainder was the previous divisor (64). Quotient = 7 (the middle quotient).

   Dividend = (128 × 7) + 64 = 896 + 64 = 960.

3. First step — its divisor was the previous dividend (960), and its remainder was the previous divisor (128). Quotient = 2 (the first quotient).

   Dividend = (960 × 2) + 128 = 1920 + 128 = 2048.

The two original numbers are the first step's divisor and dividend: 960 and 2048.

Sum = 960 + 2048 = 3008.

Cross-check

Run the forward Euclidean algorithm on 2048 and 960 to confirm: 2048 = 960×2 + 128; 960 = 128×7 + 64; 128 = 64×2 + 0. The quotients come out 2, 7, 2 and the last divisor is 64 — exactly the given data, so the pair 960 and 2048 (sum 3008) is correct.