PE 2 - LCM & HCF | Numbers

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).

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).

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.

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).

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.

Answer: Option A

Explanation :

(y² - 5y + 6) and (y² - 7y + 10)
 
y² - 5y + 6 = (y² - 3y - 2y + 6) = (y² - 3y) - (2y - 6) = y(y - 3) - 2(y - 3) = (y - 3) (y - 2)
 
y² – y - 2 = (y² - 2y + y - 2) = (y² - 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).