if a smallest no. divided by 26and32 and leaves remainder 6and12 respectivly is
Answers
396 divided by 26 and32 and leaves remainder 6 and 12 respectively
Given : A smallest no. divided by 26 and32 and leaves remainder 6 and 12 respectively
To Find : Smallest Number
Solution:
Number is N
N = 26A + 6 = 26A + 26 - 20 = 26(A + 1) - 20
N = 32B + 12 = 32B + 32 - 20 = 32(B + 1) - 20
=> N + 20 = 26(A + 1) = 32(B + 1)
Hence N + 20 is LCM of 26 and 32
"LCM - Least common multiplier of given numbers is the least number which is perfectly divisible by given numbers."
LCM = product of each factor of highest power
26 = 2 * 13
32 = 2 * 2* 2 * 2 * 2
LCM = (2 * 2 * 2 * 2 * 2 * 13) = 416
N +20 = 416
=> N = 396
396 is the required Smallest number
396 divided by 26 and32 and leaves remainder 6 and 12 respectively
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Answer:
Number is N
N = 26A + 6 = 26A + 26 - 20 20 = 26(A + 1) -20
N = 32B + 12 = 32B + 32 - 20 = 32(B + 1) - 20 => N + 20
=26(A + 1) 32(B + 1)
Hence N + 20 is LCM of 26 and 32
"LCM Least common multiplier of given numbers is the least number which is perfectly divisible by given numbers."
LCM = product of each factor of highest power
26 = 2 * 13
32 = 2*2*2*2*2
LCM = (2 * 2 * 2*2*2 *13) = 416
N + 20 = 416
=> N = 396
396 is the required Smallest number.
396 divided by 26 and32 and leaves remainder 6 and 12 respectively.