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A rectangular surface has length 4661 m and breadth 3318 m. In this area, square tiles are to be put. The maximum length of such tiles is ____ m.

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Detailed Solution

A rectangular surface has length 4661 meters and breadth 3318 meters. In this area, square tiles are to be put. The maximum length of such tiles is 79 m.
Given, a rectangular surface has length 4661 m and breadth 3318 m.
Applying Euclid’s division Lemma,

H e r e, 4661 > 3318 L e t a = 4661 a n d b = 3318 4661 = 3318 × 1 + 1343 H e r e, r = 1343 ≠ 0.

Dividend = 3318 and divisor = 1343.
Applying Euclid’s division Lemma,

H e r e, 3318 > 1343 L e t a = 3318 a n d b = 1343 3318 = 1343 ​ ​ × ​ ​ 2 + 632 H e r e, r = 632 ≠ 0

Dividend = 1343 and divisor = 632.
Applying Euclid’s division Lemma,

H e r e, 1343 > 632 L e t a = 1343 a n d b = 632 1343 = 632 ​ ​ × ​ ​ 2 + 79 H e r e, r = 79 ≠ 0

Dividend = 632 and divisor = 79.
Applying Euclid’s division Lemma,

H e r e, 632 > 79 L e t a = 632 a n d b = 79 632 = 79 × 8 + 0 H e r e, r = 0

The final value is 79 as remainder is 0.
Hence, the maximum length is 79 m.

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