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Find the least no.which when divided by 35,56 and 91 leaves the remainder 7 in each case |
| Answer» 35 = 5 {tex}\\times{/tex}\xa0756 = 23{tex}\\times{/tex}\xa0791 = 13 {tex}\\times{/tex}\xa07L.C.M of 35, 56 and 91 = 23{tex}\\times{/tex}\xa07\xa0{tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa013 = 3640The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.Hence 3647 is the smallest number that, when divided by 35, 56 and 91 leaves a remainder of 7 in each case. | |