Question

find the largest number that divides 398, 436 and 542 leaving the reminders 7, 11, 15 respectively
(please write the answer in a paper sequences and please send me the pic)
maths class 10
ch1 cbse

Answer 1

Solution : (I don't have camera to post as an image)

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Given :

To find the largest number that divides 398, 436 and 542 leaving the reminders 7, 11, 15 respectively

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We know that,

The euclid division algorithm  :

⇒ a = bq + r  (Where a > b > r ≥ 0)

So,

We can say that,

⇒ a - r = bq

Hence,

i) 398 = $bq_1$ + 7

⇒ 398 - 7 = $bq_1$

⇒ 391 = $bq_1$ ...(i)


ii) 436 = $bq_2$ + 11

⇒ 436 - 11 = $bq_2$

⇒ 425 = $bq_2$ ....(ii)


iii) 542 = $bq_3$ +15

⇒ 542 - 15 = $bq_3$

⇒ 527 = $bq_3$ ....(iii)

For all the 3 equations, value of b is same, as the divisor is same,.

Hence, we can find the number by finding HCF (As, it is given to find the largest integer),.

⇒ HCF{391,425,527} is the required number,..

⇒ HCF = product of common factors to the least power,.

⇒ 391 = 17 x 23

⇒ 425 = 5² x 17

⇒ 527 = 17 x 31
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∴ Common factor = 17
∴ Least power = 1  (As, 17 is a factor in a the 3 given numbers),.

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                  ∴  HCF{391,425,527} = 17

                  ∴ The largest integer which that divides 398, 436 and 542 leaving

the reminders 7, 11, 15 is 17,.
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                                                Hope it Helps !!

Answer 2

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.