what is the smallest number that, when divided by 35,56 and 91 leaves remainder of 5 in each case
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Find the L.C.M of 35,56,91
35 = 5*7
56 = 2^3 * 7
91 = 7*13
L.C.M = 2^3 * 5 * 7 * 13 = 8*5*7*13= 40*91 = 3640
TO get a remainder '5' just add '5' to the L.C.M== 3640+5 = 3645
Hence 3645 is the smallest number which when divided by 35,56 and 91 leave a remainder 5 in each case.
35 = 5*7
56 = 2^3 * 7
91 = 7*13
L.C.M = 2^3 * 5 * 7 * 13 = 8*5*7*13= 40*91 = 3640
TO get a remainder '5' just add '5' to the L.C.M== 3640+5 = 3645
Hence 3645 is the smallest number which when divided by 35,56 and 91 leave a remainder 5 in each case.
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