Question

the least number which when divided by 35 , 56 and 91 leaves remainder 7 in each case

Answer 1

LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640.
The smallest number which divides 35,56 and 91 is 3640. But we have a condition that it leaves a remainder of 7. So we add 7 to the LCM (3640+7=3647).

Answer 2

Heya !!!


Prime factorisation of 35 = 3 × 7


Prime factorisation of 56 = 2 × 2 × 2 × 7


And,


Prime factorisation of 91 = 7 × 13




LCM of 35 , 56 and 91 = 7 × 5 × 8 × 13 = 3640 .


Required number = LCM of 35,56 91 + 7 = 3640 + 7 = 3647.



★ HOPE IT WILL HELP YOU ★