# PE 2 - LCM & HCF | Numbers

**PE 2 - LCM & HCF | Numbers**

What is the least number that must be subtracted from 1850, so that the remainder when divided by 7, 12, and 16 is 4?

- A.
131

- B.
1355

- C.
134

- D.
166

Answer: Option D

**Explanation** :

When a number N is divided by 7, 12 and 16, the remainder is 4.

⇒ N = k × LCM (7, 12, 16) + 4, where k = 1, 2, 3… and so on.

⇒ N = 336k + 4

When k = 5, N = 1684.

So, the number to be subtracted from 1850 is = 1850 – 1684 = 166

Hence, option (d).

Workspace:

**PE 2 - LCM & HCF | Numbers**

What is the smallest number which when divided by 7, 9, and 11 leaves remainders 5, 7, and 9 respectively?

- A.
1265

- B.
691

- C.
541

- D.
1013

- E.
987

Answer: Option B

**Explanation** :

Let the required number be x.

Here divisors are 7, 9, 11 and remainders are 5, 7, 9.

The difference between respective divisors and remainders is 2.

∴ x = k × LCM (7, 9, 11) – 2 = 693k – 2.

x will be smallest when k = 1.

⇒ smallest value of x = 693 × 1 – 2 = 691.

Hence, option (b).

Workspace:

**PE 2 - LCM & HCF | Numbers**

A number leaves a remainder of 2 and 4 respectively when divided by 4 and 5. Find the least three digit such number.

Answer: 114

**Explanation** :

Let the number be N.

∴ N = 4a + 2 i.e., N can be 2, 6, 10, **14**, 18, ….

Also, N = 5b + 4 i.e., N can be 4, 9, **14**, 19, …

The least possible value of N = 14.

∴ N = k × LCM(4, 5) + 14 = 20k + 14.

Now we have to find the least three-digit such number.

For k = 5, N = 114 (least three-digit such number).

Hence, 114.

Workspace:

**PE 2 - LCM & HCF | Numbers**

Find the greatest number which when divides 252, 1848, and 462 leaves the same remainder.

- A.
64

- B.
42

- C.
36

- D.
21

- E.
84

Answer: Option B

**Explanation** :

The greatest number which when divides a, b, and c leaves the same remainder = HCF [|a – b|, |b – c|].

The greatest number which when divides 252, 1848, and 462 leaves the same remainder

= HCF ((1848 – 252), (1848 – 462)) = 42.

Hence, option (b).

Workspace:

**PE 2 - LCM & HCF | Numbers**

Find the least number which when divided by 6, 15, 17 leaves a remainder 1, but when divided by 7 leaves no remainder.

Answer: 511

**Explanation** :

Any number, which when divided by 6, 15 and 17 leaves a remainder of 1, is of the form {k × LCM(6, 15, 17) + 1}, where = 1, 2, 3… and so on.

k × LCM(6, 15, 17) + 1 = 510k + 1.

It is given that (510k + 1) is a multiple of 7.

When k = 1, we get 511, which is a multiple of 7.

Hence, 511.

Workspace:

**PE 2 - LCM & HCF | Numbers**

Find the HCF and LCM of the polynomials (y^{2} - 5y + 6) and (y^{2} - y - 2).

- A.
(y - 2), (y - 2) (y - 3) (y + 1)

- B.
(y - 2), (y - 2) (y - 3)

- C.
(y - 3), (y - 2) (y - 3) (y - 1)

- D.
(y – 3), (y + 2) (y - 3) (y - 1)

- E.
(y - 2), (y - 2)

Answer: Option A

**Explanation** :

(y^{2} - 5y + 6) and (y^{2} - 7y + 10)

y^{2} - 5y + 6 = (y^{2} - 3y - 2y + 6) = (y^{2} - 3y) - (2y - 6) = y(y - 3) - 2(y - 3) = **(y - 3) (y - 2)**

y^{2} – y - 2 = (y^{2} - 2y + y - 2) = (y^{2} - 2y) + (y - 2) = y(y - 2) + (y - 2) = **(y + 1)(y - 2)**

LCM = (y - 2) (y - 3) (y + 1)

HCF = (y - 2)

Hence, option (a).

Workspace:

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