What is the greatest number which when divides 398 and 436 leaves 7 and 11 respectively as remainder?

CBSE 10 - Maths

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How to solve this

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Asked by janhviwagh62.9dgatl | 28 Apr, 2022, 04:15: PM

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Answer

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Hint- Here, we will proceed by subtracting the given remainders from the given numbers and then, finding out the highest common factor (HCF) of the three numbers will be obtained from the last step.

Complete step-by-step answer:

The given numbers are 398, 436 and 542 which when divided by a number (largest) leaves remainders as 7, 11 and 15 respectively.So, for the numbers which will be exactly divisible by the largest number (which needs to be found), we will subtract the remainders 7, 11 and 15 from 398, 436 and 542 respectively.(398 – 7) = 391, (436 – 11) = 425 and (542 – 15) = 527According to prime factorisation, the numbers 391, 425 and 527 can be represented as the product of their prime factors as under$ 391 = 17 \times 23 \\ 425 = 5 \times 5 \times 17 \\ 527 = 17 \times 31 \\ $Clearly, we can see that 17 is the only prime factor which is common to all the numbers i.e., 391, 425 and 527.The highest common factor (HCF) of the numbers 391, 425 and 527 will be given by taking the product of all the common prime factors. Here, there is only one common prime factor that’s why the HCF will be equal to that common prime factor.HCF of 391, 425 and 527 = 17This number i.e., 17 (this is the largest number) will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectivelyTherefore, the largest required number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively is 17.Note- In this particular problem, the numbers 391, 425 and 527 will be exactly divisible by the number 17 because 17 is a prime factor (or prime number) which when divided by these numbers i.e., 391, 425 and 527 leaves zero as the remainder since, the remainders which this number 17 leaves when divided by 398, 436 and 542 are already subtracted.