find the largest number which divides 398,436 and 542 leaving remainder 7, 11 and 15 respectively
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Hence the largest number that will divide 398, 436 and 542 leaving remainder 7, 11 and 15 respectively is 17
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Hey
Here is your answer,
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.
Hope it helps you!
Here is your answer,
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.
Hope it helps you!
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