Find the smallest number when increased by 17
WebAs we know that LCM (520,468) is the least possible number which is exactly divisible by the given numbers. LCM (520,468) is 4680. So, x + 17 = 4680. ⇒ x = 4680 - 17. ⇒ x = 4663. … WebJun 30, 2024 · Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$ 3 Find the smallest number which leaves remainder 1, 2 and 3 when divided by 11, 51 and 91
Find the smallest number when increased by 17
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WebMar 25, 2024 · So, we first try to find the smallest number divisible by both $ 520 $ and $ 468 $ by taking the least common multiple of both the numbers. Then, we find our required answer by subtracting $ 17 $ from the LCM of $ 520 $ and $ 468 $ as the number has to be increased by $ 17 $ to be the LCM of the give numbers. Complete step-by-step answer: WebSep 25, 2013 · LCM of 36, 63 and 108 is 756. given that the smallest number when increased by 6 is divisible by the given numbers. Hence the smallest number is (756 –6) = 750. Recommend (3) (0) person. Prisha. take HCF of 36, 63 and 108 = 1134. now subtract 6 from 1134 = 1134 - 6 = 1128. therefor ans = 1128.
WebSolution Dear student, The given numbers are 520 and 468. The smallest number which when increased by 17 is exactly divisible by both 520 and 468 is obtained by subtracting … WebFind the smallest number which, on being decreased by 3, is completely divisible by 18, 36, 32 and 27. asked Sep 22, 2024 in Mathematics by Richa ( 61.0k points) hcf
WebMay 3, 2016 · Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468 See answers can it be negative? Advertisement Advertisement … WebMar 11, 2024 · Hey Smart Learners So in this video we are going to solve this question👉🏻 Find the smallest number which when increased by 17 is exactly divisible by both ...
WebAug 29, 2024 · Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.#gradebooster #rdsharma #hcflcm #realnumbers
WebFree various types of educational resources for kids math, what is the smallest number, find smallest number, maths numbers, number in math, math for kids, maths for kids. clock in exception formWebDec 14, 2024 · Shortcut Trick. To find the smallest number which will leave the remainder ‘11’ when divided by 516, 24, 30, and 48. L.C.M of 16, 24, 30, and 48 = 240. Smallest number = 240 × 2 + 11. = 491. Hence, “491” is the correct answer. Download Solution PDF. bocc career siteWebJun 8, 2024 · The smallest number which when increased by 7 is exactly divisible by 6 and 32 is 89. Step-by-step explanation: To find: The smallest number which both 32 and 6 can divide. To find such number we have to first find the LCM of the two numbers that are given: LCM of the number (6, 32) = 96. The number to which if 7 is added 96 bocca westonWebApr 26, 2024 · 1] Find the smallest number which when increased by 17 is exactly divisible by 520 and 468.2] Find the smallest number which when increased by 20 is exactly ... boc cccWebMar 30, 2024 · We have to find the smallest number which when increased by $17$ is divisible by two numbers. So, the required number is decreased by $17$ from $4680$. Thus, the required number is $4680 - 17 = 4663$. Note: The LCM (lowest common multiples) of the given two numbers can also be calculated by division method which is a … bocc calvert countyWebFind the smallest number which when increased by 20 is exactly divisible by 90 and 144.Part 32 of Real No. Class-10th(CBSE)same type question link :-Part 33... clock in eyeWebMar 11, 2024 · Smallest number which when increased by 11 is exactly divisible by 15, 20 and 54 = Smallest number which when increased by 11 is exactly divisible by 15, 20 and 54 = 529. Explanation: Find the LCM of 15, 20 and 54. Resolve 15,20and 54 as product prime: 15 = 3×5. 20= 2×2×5=2²×5. 54 = 2×3×3×3=2×3³. Therefore, LCM(15,20,54) = … boc cccd