Find the largest number that will divides 398 436 542 leaving remainder 7 11 and15
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First deduct the remainders from each term
398 - 7 =391
436–11 = 425
542–15 =527
Next find the HCF of the terms that you get
391 = 17x23
425 = 5x5x17
527 = 17x31
The HCF is 17.
The answer is the HCF or 17.
Check 398/17 = 23 + 7 remainder
436/17 = 25 + 11 remainder
542/17 = 31 + 15 remainder
398 - 7 =391
436–11 = 425
542–15 =527
Next find the HCF of the terms that you get
391 = 17x23
425 = 5x5x17
527 = 17x31
The HCF is 17.
The answer is the HCF or 17.
Check 398/17 = 23 + 7 remainder
436/17 = 25 + 11 remainder
542/17 = 31 + 15 remainder
abhay1951:
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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.
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