Question

What is thelargest number that divides 398, 436, and 542, leaving remainder 7, 11 and 15 respectively?

If HCF and LCM of two numbers are respectively (x-1) and (x2-1)(x2-4), then product of the two numbers will be:


a. (x2-1)(x2-4) b. (x2+1)(x2-4)

c. (x-1)2 (x2-4)(x+1)d. (x2-1)(x2+1)(x-4)

3. Which is the least number that is divisible by all the numbers from 1 to 10 (both inclusive)?

4. Cube of any positive integer of the form 3m+2 is of the form:

a. 3q+1 b. 3q-2 c. 9q+8d. none of these

** IF POSSIBLE, PLEASE LET ME KNOW THE METHOD FOR SOLVING THESE QUESTIONS.

Answer 1

option c is right because product of two numbers equal to their LCM and HCF product

Answer 2

Given :-

398 , 436 and 542

To Find :-

The largest number

Solution :-

Let’s assume the integer is x

According to the condition given in the question

⇒ xy+7 = 398

⇒  xz+11 = 436

⇒  xk+15 = 542

⇒ xy =391

⇒  xz = 425

⇒  xk = 527

⇒  17 × 23 = 391

⇒  17 × 25 = 425

⇒ 17 × 31 = 527

So, the largest possible integer that will divide 398,436,542 & leaves reminder 7,11 and 15 respectively was 17.