Question

By using Euclid Division algorithm, find the largest number which when divides 969 and 2059 gives the remainders 9 and 11, respectively.

Answer should more than 45+ words☑️​​

Answer 1

$\huge\pink{\mid{\fbox{\tt{Answer}}\mid}}$

⇒ 64

$\huge\purple{\mid{\fbox{\tt{Solution}}\mid}}$

$\tt \red {GIVEN : }$

Numbers which are given are 969 and 2069 and theirs remainders 9 and 11 respectively.

$\tt \green{TO \: FIND :}$

The largest number which when divides 969 and 2059 gives the remainders 9 and 11, respectively by using the Euclid's division lemma.

$\tt \orange {NOTE : }$

First, we subtract the remainder from the given numbers and then calculate the HCF of new numbers

$\tt \blue{CALCULATIONS :}$

Given numbers are 969,2069 and remainders are 9,11 respectively. Then, new numbers after subtracting remainders are :-

$\tt \implies \: 969 - 9 = 960$

$\tt \: and \: 2059 - 11 = 2048$

Here, 2048 > 960

Now, by using Euclid's division lemma, we get :-

$\tt \red{\implies \: 2048 = (960 × 2) + 128}$

Here, remainder = 128

So, on taking 960 as the new dividend and 128 as new divisor and then apply Euclid's division lemma , we get :-

$\tt \pink{\implies \: 960 = (128 × 7) + 64 }$

Again, remainder = 64

So, on taking 128 as new dividend and 64 as new divisor and then apply Euclid's division lemma, we get :-

$\tt \purple{\implies \: 128 = (64 × 2) + 0}$

Here, remainder = 0

Since, the remainder has now become zero and the last divisor is 64

∴ HCF of 2048 and 960 is 64.

Hence, the required largest number is 64.

Answer 2

Answer:

GIVEN:

Numbers which are given are 969 and 2069 and theirs remainders 9 and 11 respectively.

TO FIND:

The largest number which when divides 969 and 2059 gives the remainders 9 and 11, respectively by using the Euclid's division lemma.

NOTE:

First, we subtract the remainder from the given numbers and then calculate the HCF of new numbers

CALCULATIONS:

Given numbers are 969,2069 and remainders are 9,11 respectively. Then, new numbers after subtracting remainders are :-

969 - 9 = 960⟹969−9=960

2059 - 11 = 2048 and2059−11=2048

Here, 2048 > 960

Now, by using Euclid's division lemma, we get :-

2048 = (960 × 2) + 128}⟹2048=(960×2)+128

Here, remainder = 128

So, on taking 960 as the new dividend and 128 as new divisor and then apply Euclid's division lemma , we get :-

960 = (128 × 7) + 64 }⟹960=(128×7)+64

Again, remainder = 64

So, on taking 128 as new dividend and 64 as new divisor and then apply Euclid's division lemma, we get :-

128 = (64 × 2) + 0}⟹128=(64×2)+0

Here, remainder = 0

Since, the remainder has now become zero and the last divisor is 64

∴ HCF of 2048 and 960 is 64.

Hence, the required largest number is 64.