Question

find the least number which after being increased by 5 is exactly being divisible by 7,12 and 14​

Answer 1

Answer:

First find the least number exactly divisible by 7,12 and 14

12= 2*2*3

14=7*2

So, least number = 7*2*2*3 = 84

So, least number which when increased by 5 is divisble by 7,12,14is 84-5 =79

Step-by-step explanation:

Answer 2

Let the number divisible by 7, 12 and 14 be 'x' and the number which is originally increased to obtain x be 'a'

hence, x = a + 5 ... (i)

i.e. 'x' is the L.C.M. of 7, 12 and 14

now,

7 = 1 × 7

12 = 2 × 2 × 3

14 = 2 × 7

H.C.F. = 1

L.C.M. = H.C.F. × product of remaining factors

L.C.M. = 1 × 7 × 2 × 2 × 3 × 2 × 7

L.C.M. = 7 × 12 × 14

L.C.M. = 1176

x = 1176

a + 5 = 1176 ... (from i)

a = 1176 - 5

a = 1171

Hence, the least number which after being increased by 5 is exactly divisible by 7, 12 and 14 is 1171