Math, asked by vandhana240, 1 year ago

Find the largest number which divides 398 436 and 542 leaving remainder 7 11 and 15 respectively

Answers

Answered by Anonymous
842

Answer :-

→ 17 .

Step-by-step explanation :-

We have ,

The number 398, 436 and 542 which when divides by a positive integers leaves remainder as 7, 11, and 15 respectively .

Clearly, the required number divides ( 398 - 7 ) = 391 , ( 436 - 11 ) = 425, and ( 542 - 15 ) = 527 exactly .

•°• Required number = HCF( 391, 425, 527 ) .

Now,

→ 391 = 17 × 23 ,

→ 425 = 5² × 17 ,

→ 527 = 17 × 31 .

 \therefore HCF( 391, 425, 527 ) = 17 .

Hence, the required number is 17 .

Answered by durgeshbishi2
4

Answer: The required number is 17.

Step-by-step explanation:

On dividing 398 by the required number, there is a remainder of 7.

This means that 398 7 = 391 is exactly divisible by the required number. Similarly, 436 -11 = 425 and 542-15 = 527 are exactly divisible by the required number.

The HCF of two positive integers is the largest positive integer that divides both the integers.

So, the required number will be the HCF of 391, 425, and 527. And that can be found by using Euclid's division algorithm.

425 = 391 x 1 + 34

391 = 34 x 11 + 17

34 = 17 x 2 + 0

Thus, HCF = 17

Hence, the required number is 17.

#SPJ2

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