Find the least number which when divided by 35 56 and 91 leaving the same remainder 7 in each case

Find the least number which when divided by 35 56 and 91 leaving the s

1.	Find the least number which when divided by 35 56 and 91 leaving the same remainder 7 in each case
Answer» The smallest number which when divided by 35, 56 and 91 =\xa0LCM(35,56,91)35=5×756=23×791=7×13LCM=23×5×7×13=3640The\xa0smallest\xa0number\xa0that\xa0when\xa0divided\xa0by\xa035,\xa056,\xa091\xa0leaves\xa0a\xa0remainder\xa07\xa0in\xa0each\xa0case\xa0=\xa03640\xa0+\xa07\xa0=\xa03647. Least number which can be divided by 35, 56 and 91 is LCM of 35, 56 and 91. Prime factorization of 35, 56 and 91 is: 35 = 5 × 7\xa0& 56 = 23 × 7\xa0& 91 = 7 × 13 LCM =23 × 5 × 7 × 13 = 3640 Least number which can be divided by 35, 56 and 91 is 3640. Least number which when divided by 35, 56 and 91 leaves the same remainder 7 is 3640 + 7 = 3647.