Find the largest number which divides 245 and 1029 leaving remainder 5 in each case.

hello guys is video mein Hamen largest number find karna hai jo 245 1029 Ko Jab divide Kare 25 reminder chhode to Kahane Ka Matlab Ki 5 reminder Chhod raha hai to Agaram 5 ko Hata De Tu Jo Bacha Hua number hai usko divide karna chahie to Kahane ka matlab Hamen HCF chahie to 45 - 52 Patti aur 1029 minus 5 is equal to 1024 X 1024 X 1024 210 aur 242 into 2 into 2 into 2 into 15 likhate Ho To Ho Jaega 2 ki power 4 Tujhse kya dekh sakte hain highest common factor aaega to keep our Core DG Ne 5 is -2 Kyunki aap 245 ko

Hint: To find the largest divisible number leaving remainder $5$ each. Find the required number subtracting $5$ from each. And then find H.C.F. of obtained numbers and the result we get. Complete step-by-step answer: Subtract 5 from each number we get the required number. The required number divides $\left( \right) = 24$ Hence, highest common factor of required number is $ = 24$ Therefore, the required number is also $24$ Note: The highest common factor is found by multiplying all factors which appear in both lists of factors of given two numbers. In this problem it is required to find the largest number which divides two given number but leaving a remainder, in this case to find exact divisible number i.e. there should not be any remainder, we need to subtract remainder from number so that no remainder is left the number is divisible .Hence we have subtract $5$ from each. Similarly if we need to find the smallest number divisible by a given number we will find the lowest common factor. Common factors play a vital role in finding divisible numbers.

Ans. Since we want the remainder to be 5 , therefore the required number is the HCF of $(245-5)$ and $(1029-5)$ i.e., 240 and 1024 Step 1: Since1024>240 , we will apply Euclid’s division lemma such that $1024=240 \times 4+64$ Step 2: Since the remainder in (1) is not zero so we have to apply the Euclid’s division lemma to 240 and 64, such that $240=64 \times 3+48 \ldots \ldots(2)$ Step 3: Again, the remainder in $(2)$ is not zero so we have to apply the Euclid’s division lemma to 64 and 48, such that $64=48 \times 1+16, \ldots(3)$ Step $4:$ Again, the remainder in (3) is not zero so we have to apply the Euclid’s division lemma to 48 and 16, such that $48=16 \times 3+0$ Now the remainder in equation (4) is zero. Therefore, HCF of 1024 and 240 is ’16’. The largest number which divides 245 and 1029 leaving remainder 5 in each case is $16 .$ (i) $2 \mathrm$ GJ and equilibrium constant of the reactions.

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