what is the smallest no. that when divided by 35,56,91 leaves a remainder 7 in each case?

The smallest number which when divided by 35, 56 and 91 = LCM of 35, 56 and 91

35 = 5 x 7

56 = 2 x 2 x 2 x 7

91 = 7 x 13

LCM = 7 x 5 x 2 x 2 x 2 x 13 = 3640

The smallest number that when divided by 35, 56, 91 leaves a remainder 7 in each case = 3640 + 7 = 3647.

- The number which is exactly divisible by 8, 15 and 21 = LCM of 8, 15 and 21.

8 = 2 x 2 x 2

15 = 3 x 5

21 = 3 x 7

LCM of 8, 15 and 21 = 2 x 2 x 2 x 3 x 5 x 7 = 840

110000 รท 840 = 13 R 800

[The quotient when 110000 is divided by 840 is 13 and the remainder is 800.]

Therefore, the number whixh is nearer to 110000 and greater than 100000 which is exactly divisible by 8, 15 and 21 = 110000 - 800 = 109200.

- Circumference of a circular field = 360 km.

Three cyclists an cycle 48 km, 60 km and 72 km in a day.

LCM of 48, 60 and 72.

48 = 2 x 2 x 2 x 2 x 3

60 = 2 x 2 x 3 x 5

72 = 2 x 2 x 2 x 3 x 3

LCM of 48, 60 and 72 = 2 x 2 x 2 x 2 x 3 x 3 x 5

= 720.

If the cyclists start together, they meet after 720 km i.e. after two rounds.