Question

Divide 13/80 by 5/16​

Answer 1

Let x be any positive integer and y = 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = m

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 …. (2)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 …. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 …. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 …. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,x2= 3m + 1 (3)

Let x be any positive integer and y = 3.By Euclid’s division algorithm, then,x = 3q + r for some integer q≥0 and r = 0, 1, 2, as r ≥ 0 and r < 3.Therefore, x = 3q, 3q+1 and 3q+2Now as per the question given, by squaring both the sides, we get,x2 = (3q)2 = 9q2 = 3 × 3q2Let 3q2 = mTherefore, x2= 3m ……………………..(1)x2 = (3q + 1)2 = (3q)2+12+2×3q×1 = 9q2 + 1 +6q = 3(3q2+2q) +1Substitute, 3q2+2q = m, to get,x2= 3m + 1 …. (2)x2= (3q + 2)2 = (3q)2+22+2×3q×2 = 9q2 + 4 + 12q = 3 (3q2 + 4q + 1)+1Again, substitute, 3q2+4q+1 = m, to get,x2= 3m + 1 (3)Hence, from equation 1, 2 and 3, we can say that, the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Answer 2

We know that if the denominator of a rational number has no prime factors other than 2 or 5, then it is expressible as a terminating, otherwise it has non - terminating repeating decimal representation. Thus, we will have to check the prime factors of the denominators of each of the given rational numbers.

In \frac{13}{80}

80

13

, the denominator is 80.

We have, 80 = 2 × 2 × 2 × 2 × 5.

Thus, 80 has only 2 and 5 as prime factors.

Hence, \frac{13}{80}

80

13

must have a terminating decimal representation.

∴ Terminating decimal representation of \frac{13}{80}

80

13

= 0.1625

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