Find the least no which when divided
WebSo, let to find the least number of 5 digits divisible by given numbers is x ⇒x>9999 360x > 3609999= 36999.9=27.775 x is completely divisible by 360 So, 360x is an integer. So, we need to find next integer greater than 27.775 So, completely divisible 5 digit number by 360 is 360×28=10080 Weba. Find the least positive integer with remainders 1,2, and 3 when divided by 7,8, and 9 respectively. The three congruences are $ x \equiv 1 $ (mod 7) $ x \equiv 2 $ (mod 8) $ x \equiv 3 $ (mod 9)
Find the least no which when divided
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WebFeb 3, 2024 · The least number which when divided by 4, 6, 8, 12 and 16 leaves a remainder of 2 in each case is: ... The exam is held to find candidates for the positions of Senior Assistant, Junior Assistant, and Junior Accountant. Last year, 42 NBE Recruitment positions were made available. This year, the board is expected to fill the same or more ... WebAug 20, 2016 · Remark $ $ Only very small numbers were involved because we solved the congruences with largest moduli first, so that, in the end, when the numbers are bigger, this is compensated by computing at smaller moduli. We simplify arithmetic by using remainders of least magnitude, e.g. we used $\,-4\equiv 31\pmod{35}$. Note that we did not not …
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WebFind the least number which when divided by 20, 25, 35 and 40 leaves remainders 14, 19, 29 and 34 respectively. Solution First find out the LCM of 20,25,35 and 40 20=2×2×5 … WebFeb 23, 2024 · ∴ The least number which leaves remainder 3 and 7 respectively when divided by 7 and 11 is 73. Alternative Method, Given, Divisors = 7 and 11 The LCM of 7 and 11 = 77 Now, 7 - 3 = 4 and 11 - 7 = 4 By subtracting 4 from 77, we get the number = 77 - 4 = 73 ∴ The least number which leaves remainder 3 and 7 respectively when divided by …
WebFind the least number which when divided by 25, 40 and 60 leaves 9 as the remainder in each GRAVITY COACHING CENTRE 40K views 4 years ago LCM problem where remainder is not same and difference...
WebFind the smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively. Medium Solution Verified by Toppr The required number is determine with … image fire extinguisherimage firebaseWebApr 5, 2024 · By substituting the value of n in above equation, we get: Required number = (60 x 5) + 1 = 301 301 is divisible by 7. Therefore 301 is the least number which leaves remainder as 1 when divided by 2, 3, 4, 5, 6 and leaves no remainder when divided by 7. Option (b) is correct. image fire forceWebTherefore, 3647 is the smallest number that when divided by 35, 56 and 91 leaves remainder of 7 in each case. ... Find the least number that is divisible by all the numbers between 1 and 10 (both inclusive). In a school there are two sections – section A and section B of Class VI. ... image first aid boxWebJan 17, 2024 · First, if a number is being divided by 10, then the remainder is just the last digit of that number. Similarly, if a number is being divided by 9, add each of the digits to each other until you are left with one number (e.g., 1164 becomes 12 which in turn becomes 3), which is the remainder. image firstWebWe have to determine the least number which when divided by 2, 3, 4, 5 and 6 leaves 1 as a remainder and no remainder when divided by 7 Solution Let us determine the L C M of the numbers 2, 3, 4, 5 and 6 The value of L C M ( 2, 3, 4, 5, 6) obtained will be the smallest number that is exactly divisible by 2, 3, 4, 5 and 6 image fire and brimstoneWebWhen the same number is divided by 24, it leaves a remainder of 13. Observing the constant difference of 11 between divisor and remainder: We can conclude that if 11 is subtracted from the LCM of 18, 21 and 24, then it will leave a remainder of 7 when divided by 18, a remainder of 10 when divided by 21 and a remainder of 13 when divided by 24. image firefighter