What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3
What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively? Show
Answer  Hint: For solving this question you should know about the Greatest Common Factor between the two or more digits. For calculating the greatest common factor or GCF, we will do factors of the given digits and then we will take all the common values and see the greatest common value in that. But before that, we will subtract the remainders of that and then we find some values and these values will give us the greatest common factor or the largest number which will divide them. Complete
step-by-step solution: Note: During solving this question you can use any one of the methods, but the first method is more tough and that has chances of mistakes in the answer because factors can be wrong and by mistake, we can forget to take the common one from the value of the factors, in which case our answer would be wrong. The required number when divides 626, 3127 and 15628, leaves remainder 1, 2 and 3. This means 626 – 1 = 625, 3127 – 2 = 3125 and 15628 – 3 = 15625 are completely divisible by the number ∴ The required number = HCF of 625, 3125 and 15625 First consider 625 and 3125 By applying Euclid’s division lemma 3125 = 625 × 5 + 0 HCF of 625 and 3125 = 625 Now consider 625 and 15625 By applying Euclid’s division lemma 15625 = 625 × 25 + 0 ∴ HCF of 625, 3125 and 15625 = 625 Hence required number is 625 the question is that what is the largest number that divides 626 3027 and 15006 28 and leaves remainders of 1 2 and 3 respectively so here 626 when divided leaves remainder 1 so it will be 26 - 16 25 S 13127 will leave reminder to so it will be 3125 after subtracting and the third one will be 15628 that will be 15625 so the required number is HCF of 6 25312 5 and 15 625 now using euclid's division Lemma 3125 can be written as 625 in 25 + 0 and 15625 may be written as 625 into 25 + zero so clearly since we are getting zero as remainder so we may say that HCF is 625 our final answer will be the required just number is 625 I hope you understood explanation thank you What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively. Clearly the required number is the HCF of the following numbers 626 - 1 = 625, 3127 - 2 = 3125 and 15628 - 3 = 15625 Case I. Finding the HCF of 625 and 3125 by applying Euclid’s division lemma. I. 3125 = 625 × 5 + 0 Since, the remainder at this stage is zero, so the divisor i.e., 625 at this stage is the HCF of 625 and 3125. Case II. Finding the HCF of 625 and third number 15625 by applying Euclid’s division lemma.
Hence, HCF of (626, 3127, 15628) is 625. 1236 Views Express each number as a product of its prime factors: (i) 140
(i) 140 So, 140 = 2 × 2 × 5 × 7 = 22 × 5 × 7. 548 Views Use Euclid’s division algorithm to find the HCF of: (i)135 and 225 (ii) 196 and 38220. (i) Given integers are 135 and 225, clearly 225 > 135. Therefore, by applying Euclid’s division lemma to 225 and 135, we get If. Since, the remaindei 90 ≠ 0, we apply division lemma to 135 and 90, to ge. We consider the new divisor 90 and new remainder 45 and apply division lemma to get The remainder of this step is zero. So, the divisor at this stage or the remainder at the previous stage i.e., 45 is the HCF of 135 and 225. (ii) Given integers are 196 and 38220. Therefore by applying Euclid’s division lemma to 196 and 38220, we get The remainder at this step is zero. So. our procedures stops and divisor at this stage i.e. 196 is the HCF of 196 and 38220. 3449 Views Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. 197 Views Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Let a and b be two positive integergs such that a is greater than b; then : a = bq + r; where q and r are positive integers 0 ≤ r < b. Taking b = 3, we get a = 3q + r ; where 0 ≤ r < 3. ⇒ Different values of integer a are 3q, 3q + 1 or 3q + 2. Cube of 3q = (3q)3 = 27q3 = 9(3q3) = 9m ; where m is some integer. Cube of 3q + 1 = (3q + 1)3 = (3q + 3(3q)2 × 1 + 3(3q) × 12+ l3 [∵ (a + b)3 = a3 + 3a2b + 3ab2 + 1] = 27q3 + 27q2 + 9q + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1; where m is some integer. Cube of 3q + 2 = (3q + 2)3 = (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23 = 27q3 + 54q2 + 36q + 8 = 9(3q3 + 6q2 + 4q) + 8 = 9m + 8; where m is some integer. ∴ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8. 3180 Views Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Let a be any positive integer. Let q be the quotient and r be remainder. Then a = bq + r where q and r are also positive integers and 0 ≤ r < b Taking b = 3, we get a = 3q + r; where 0 ≤ r < 3 When, r = 0 = ⇒ a = 3q When, r = 1 = ⇒ a = 3q + 1 When, r = 2 = ⇒ a = 3q + 2 Now, we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m or 3m + 1 for some integer m. ⇒ Squares of 3q = (3q)2 = 9q2 = 3(3q)2 = 3 m where m is some integer. Square of 3q + 1 = (3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2 q) + 1 = 3m +1, where m is some integer Square of 3q + 2 = (3q + 2)2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1)+ 1 = 3m + 1 for some integer m. ∴ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m. 1156 Views Is the largest number that divides 626 3127 and 15628 and leaves remainders of 1/2 and 3 respectively?Therefore, the largest number that divides 626, 3127 and 15628 and leaves remainders 1, 2 and 3 is 625.
What is the largest number that divides 70 and 125?Hence, 13 is the largest number which divides 70 and 125, leaving remainders 5 and 8 respectively.
What is the largest number that divides 70 and 125 leaving remainders?Hence, the largest number that divides 70 and 125 , leaving remainders 5 and 8 , respectively is 13 .
What is the largest number that divide?The largest no. that divides two numbers is known as the GCD (Greatest Common Divisor) or the HCF (Highest Common Factor).
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