find the greatest number that divides 398 436 and 542 leaving reminders 711 and 13 respectively

Answer: No such number exists (the HCF of $391$ , $425$ , and $529$ is $1$ , which is not greater than the remainders $7$ , $11$ , and $13$ ).

Step-by-step solution

Step 1: Subtract the remainders from each given number

If a number divides $398$ , $436$ , and $542$ leaving remainders $7$ , $11$ , and $13$ respectively, it must divide the differences completely. Subtracting the respective remainders gives $391$ , $425$ , and $529$ . Therefore, the required number is the highest common factor (HCF) of $391$ , $425$ , and $529$ .

Step 2: Find the HCF of 391 and 425 using Euclid's division lemma

Applying Euclid's division algorithm to $425$ and $391$ , we divide $425$ by $391$ to get a remainder of $34$ . Next, dividing $391$ by $34$ yields a remainder of $17$ . Finally, $34$ divided by $17$ leaves a remainder of $0$ , which gives $\text{HCF}(391, 425) = 17$ .

Step 3: Find the HCF of 17 and the third number, 529

Now we find the HCF of the result, $17$ , and the third number, $529$ . Dividing $529$ by $17$ gives a quotient of $31$ and a non-zero remainder of $2$ . Since $17$ is a prime number and does not divide $529$ , the common factor of all three numbers is only $1$ .

Step 4: Evaluate the validity of the divisor

The mathematical $\text{HCF}(391, 425, 529) = 1$ . However, in any division, the divisor must strictly be greater than each remainder ( $d > 7$ , $d > 11$ , and $d > 13$ , so $d > 13$ ). A divisor of $1$ cannot leave remainders of $7$ , $11$ , and $13$ , which shows that no such positive integer exists for the numbers given as stated.

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