**Practice Problems Based on Division Algorithm for Grade 10 :**

Here we are going to see some practice problems based on division algorithm for grade 10.

To find questions from 1 to 4, please visit the page "Word Problems Based on Division Algorithm".

**Question 5 :**

Prove that square of any integer leaves the remainder either 0 or 1 when divided by 4.

**Solution :**

Let n be the integer, we wish to square.

**Case 1 :**

If n is even...

let n = 2m

n^{2 }= 4m^{2}

Hence n is divisible by 4 and leaves the remainder 0.

**Case 2 :**

If n is odd...

let n = 2m+1

n^{2} = (2m + 1)^{2}

n^{2} = (2m)^{2} + 2(2m) + 1^{2}

n^{2} = 4m^{2} + 4m + 1

n^{2} = 4(m^{2} + m) + 1

Hence n is divisible by 4 and leaves the remainder 1.

**Question 6 :**

Use Euclid’s Division Algorithm to find the Highest Common Factor (HCF) of

(i) 340 and 412

**Solution :**

a = 340 and b = 412

412 > 340

412 = 1 (340) + 72

340 = 4(72) + 52

72 = 1(52) + 20

52 = 2(20) + 12

20 = 1(12) + 8

12 = 1(8) + 4

8 = 2(4) + 0

Hence the required H.C.F is 4.

(ii) 867 and 255

**Solution :**

a = 867 and b = 255

867 > 255

867 = 3(255) + 102

255 = 2(102) + 51

102 = 2(51) + 0

Hence the H.C.F is 51.

(iii) 10224 and 9648

**Solution :**

10224 > 9648

10224 = 1(9648) + 576

9648 = 16(576) + 432

576 = 1(432) + 144

432 = 3(144) + 0

Hence the H.C.F is 144.

(iv) 84, 90 and 120

**Solution :**

90 > 84

90 = 1(84) + 6

84 = 6(14) + 0

Hence the required H.C.F is 6.

**Question 7 :**

Find the largest number which divides 1230 and 1926 leaving remainder 12 in each case.

**Solution :**

1230 - 12 = 1218

1926 - 12 = 1914

Now we have to find H.C.F of 1218 and 1914.

1914 = 1218(1) + 696

1218 = 696(1) + 522

696 = 522(1) + 174

522 = 174(3) + 0

Hence the largest number which divides 1230 and 1926 leaving remainder 12 is 174.

**Question 8 :**

If d is the Highest Common Factor of 32 and 60, find x and y satisfying d = 32x + 60y .

**Solution :**

60 > 32

60 = 32(1) + 28 ----(1)

32 = 28(1) + 4 ----(2)

28 = 4(7) + 0 ----(3)

Hence the H.C.F of 60 and 32 is 4.

d = 4

d = 32x + 60y

4 = 32 x + 60 y

From (2), let us solve for 4

4 = 32 - 28(1) ---(4)

From (1), let us solve for 28

28 = 60 - 32(1)---(5)

By applying the value of 28 from (5) into (4), we get

4 = 32 - (60 - 32(1)) ⋅ 1

4 = 32 - 60 ⋅ 1 + 32 (1) ⋅ 1

4 = 32 + 32 (1) - 60 ⋅ 1

4 = 32(1 + 1) - 60 ⋅ 1

4 = 32(2) + 60 (-1)

Instead of x, we have 2 and instead of y, we have -1.

Hence x = 2 and y = -1.

**Question 9 :**

A positive integer when divided by 88 gives the remainder 61. What will be the remainder when the same number is divided by 11?

**Solution :**

Let "n" be that integer number.

We are given that (n-61) is divisible by 88.

Hence, (n-61) is divisible by 11.

It is the same as to say that ((n-55)-6) is divisible by 11.

55 is divisible by 11

It means that the number n gives the remainder 6 when is divided by 11.

**Question 10 :**

Prove that two consecutive positive integers are always coprime.

**Solution :**

Let two consecutive numbers are x and x + 1

H.C.F of (x, x + 1) = n, because it cannot be equal to 1, n is natural and n > 1

n divides n both x and x + 1.

then n divides x + 1 - x

Hence n divides 1 or n = 1

But we have assumed n > 1

So by contradiction x & x + 1 are co-prime.

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