Question

Find the greatest number which exactly divides 280 and 1245 leaving remainder 4 and 5 respectively​

Answer 1

➢ Correct question :

Find the greatest number which exactly divides 280 & 1245 leaving a remainder of 4 and 3 respectively.

➢ Answer :

The required number = 138

➢ Given :

The number which leaves a remainder of 4 & 3 on dividing 280 & 1245

➢ To Find :

The required number.

➢ How To Find :

We are given that the number will leave a remainder of 4 & 3 on dividing 280 & 1245

Now If we subtract the given remainders from these numbers, the resulting numbers will be exactly divisible by our required number (leaving no remainders).

⇝ 280 - 4 = 276

⇝ 1245 - 3 = 1242

Therefore, the required number is the HCF of 276 & 1242.

By prime factorisation method we get,

⇝ 276 = 2² × 3¹ × 23¹

⇝ 1242 = 2¹× 3³ × 23¹

∴ HCF = 2¹ × 3¹ × 23¹ = 138.

Hence, the greatest number which exactly divides 280 & 1245 leaving a remainder of 4 & 3 is 138.

Answer 2

YOUR CORRECT QUESTION :

Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.

AND UA ANSWER IS.....

SOLUTION:

To find the greatest number which when divides 280 and 1245 leaving the remainders 4 and 3 respectively. First ,we subtract the remainder from the given numbers and then calculate the HCF of new numbers.

Given numbers are 280 and 1245 and remainders are 4 and 3.

Then ,new numbers after subtracting remainders are :

280 - 4 = 276 and 1245 - 3 = 1242

Now, we have to find the H.C.F. of 276 and 1242.

By applying Euclid’s division lemma,a = bq+r

Let a = 1242 and b = 276

1242 = 276 x 4 + 138

276 = 138 x 2 + 0.

Here remainder is zero , and the last divisor is 138.

So H.C.F is 138

Hence, the required greatest number is 138

HOPE THIS ANSWER WILL HELP YOU.....

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