find the greatest number which divides 398 436. and 542 leaving remainder 7 11 and 15 respectively
Answers
Answer :-
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 .
HCF( 391, 425, 527 ) = 17 .
Hence, the required number is 17 .
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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.